Calorimetry

Let a substance A of mass m₁, specific heat capacity c₁ at temperature t₁ is mixed with another substance B of mass m₂, specific heat capacity c₂ at a lower temperature t₂.

If the final temperature of the mixture becomes t, then

Fall in temperature of substance A = t₁ – t

Rise in temperature of substance B = t – t₂

Heat energy lost by A = m₁ × c₁ × fall in temperature

= m₁c₁ (t₁ – t)

Heat energy gained by B = m₂ × c₂ × rise in temperature

= m₂c₂ (t – t₂)

If no heat energy is lost in the surrounding, then by the principle of mixtures,

Heat energy lost by A = Heat energy gained by B

m₁c₁ (t₁ – t) = m₂c₂ (t – t₂)

m₁ c₁ t₁ - m₁c₁ t = m₂ c₂ t - m₂ c₂ t₂

m₁c₁ t₁ + m₂c₂ t₂ = m₁c₁ t + m₂c₂ t

m₁c₁ t₁ + m₂c₂ t₂ = t (m₁c₁ + m₂c₂)

Therefore,

$\text{t} = \dfrac{\text{m}_1 \text{c}_1 \text{t}_1 + \text{m}_2 \text{c}_2 \text{t}_2}{\text{m}_1\text{c}_1 + \text{m}_2\text{c}_2}$

The assumption made here is that there is no loss of heat energy.

The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence, the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will impart nearly five times more heat energy than that given by the same mass of sand for the same fall in temperature.

As such sand (or earth) gets heated or cooled more rapidly as compared to water under the similar conditions (exposure to the Sun). Thus, near the sea shore, there becomes a large difference in temperature between the land and sea due to which convection currents are set up. The cold air from the land blows towards the sea during the night (i.e., land breeze) and during the day cold air blows from the sea towards the land (i.e., sea breeze). These breezes near the sea makes the climate in coastal areas moderate.

(a) A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained (or lost) by a body when it is mixed with the other body.

(b) It is made up of a thin sheet of copper. The reasons for using copper are:

1. Copper is a good conductor of heat, so the vessel soon acquires the temperature of it's contents.
2. Copper has the low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from the contents to acquire it's temperature, is very small.

(c) Heat capacity of the calorimeter should be low. Hence, B should be preferred for making the calorimeter.

(d) The outer and inner surfaces of the vessel should be polished so as to reduce the loss of heat due to radiation in a calorimeter.

thermal energy

Reason — Thermal energy refers to the total internal energy of a substance arising from the random motion and interactions of its constituent molecules. This internal energy includes the kinetic energy of the molecules (their motion) as well as the potential energy associated with their intermolecular forces and interactions.

Both (2) and (3)

Reason — S.I. unit of heat is Joule (J). The other most commonly used unit of heat is Calorie (cal).

4200 J kg^-1 K^-1

Reason — Water has an unusually high specific heat capacity of 4200 J kg^-1 K^-1

flow of heat

Reason — Temperature determines the direction of heat flow because heat naturally flows from regions of higher temperature to regions of lower temperature.

all of the above

Reason — The amount of heat energy contained in a body depends on its

1. Mass — The greater the mass of a body, the more heat energy it can contain, assuming all other factors remain constant.
2. Temperature — The temperature of a body directly influences the amount of heat energy it contains. Higher temperatures indicate higher amounts of thermal energy.
3. Material of the Body — Different materials have different specific heat capacities, which determine how much heat energy is required to raise the temperature of the material by a certain amount.

C' = m x c

Reason — C'= m x c hence, the heat capacity (C') is directly proportional to both the mass (m) and the specific heat capacity (c).

low, high

Reason — The substance with low specific heat capacity shows a rapid and high rise in temperature thus it is a better conductor of heat than the substance with high specific heat capacity which shows a slow and small rise in temperature.

hydrogen

Reason — The specific heat capacities of the given elements are:

Hence, Hydrogen has the highest specific heat capacity.

both (1) and (2)

Reason — Law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed from one form to another. In calorimetry, heat energy is exchanged between a system and its surroundings, but the total amount of heat energy in the system and surroundings remains constant, in accordance with these conservation laws.

principle of calorimetry

Reason — Heat is measured by calorimetry, which involves the determination of heat changes in a system through the measurement of temperature changes.

Heat energy lost by a hot body = Heat energy gained by the cold body.

This is called the principle of calorimetry.

assertion is true but reason is false.

Explanation

Assertion (A) is true. The specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of the substance by 1°C (or 1 K).

Reason (R) is false. The specific heat capacity of a substance is its characteristic property. It represents the substance's ability to store heat energy per unit mass and is an intrinsic property that is unique to each substance.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true. The principle of the method of mixtures involves mixing substances at different temperatures to find the final temperature. This method is based on the principle that heat lost by the hotter substance is equal to the heat gained by the colder substance when they reach thermal equilibrium.

Reason (R) is true. The law of conservation of energy states that energy is neither created nor destroyed in an isolated system. This principle is fundamental in understanding heat transfer processes, including the method of mixtures. In the method of mixtures, the total energy of the system (the substances being mixed) remains constant, as energy is transferred from the hotter substance to the colder substance until thermal equilibrium is reached.

A degree (or temperature difference) is same on both the celsius and kelvin scales i.e., △t° C = △T K. Therefore, the corresponding rise in temperature on the Kelvin scale will be 15 K.

(a) Given,

mass (m) = 200 g = 0.20 kg

specific heat capacity (c) = 410 J kg^-1 K^-1

We know that,

Heat capacity (C^') = Mass (m) × specific heat capacity (c)

Substituting the values in the formula above we get,

C' = 0.20 x 410 = 82 J K^-1

Hence, heat capacity of a copper vessel = 82 J K^-1

(b) Change in temperature = 35°C - 25°C = 10°C = 10 K

Energy required to increase the temperature of vessel Q = ?

From relation,

Q = m x c x △T

Substituting the values in the formula above we get,

Q = 0.20 x 410 x 10 = 820 J

Hence, heat energy required to increase the temperature of the vessel = 820 J

(i) Given,

mass (m) = 2.0 kg

heat capacity (C') = 966 J K^-1

rise in temperature (△t) = 15°C

Heat energy needed (Q) = ?

From relation,

Q = C' × △t

Substituting the values in the formula above we get,

Q = 966 x 15 = 14,490 J

Hence, heat energy needed to warm it by 15°C = 14,490 J

(ii) specific heat capacity (c) = ?

We know that,

$c = \dfrac{C'}{m} \\[0.5em]$

Substituting the values in the formula above we get,

$c = \dfrac{966}{2} = 483 \text{ J kg}^{-1} \text{K}^{-1}$

Hence, Specific heat capacity = 483 J kg^-1 K^-1

Given,

Mass of copper (m) = 200 g = 0.2 kg

Change of temperature (△t) = (70 - 20)°C = 50°C

Specific heat capacity of copper (c) = 390 J kg^-1 K^-1

Amount of heat required to raise the temperature of 0.2 kg of copper = ?

From relation,

Q = m x c x △t

Substituting the values in the formula above we get,

Q = 0.2 x 390 x 50 = 3900 J

Hence, the amount of heat energy required to raise the temperature of 200 g of copper from 20° C to 70° C = 3900 J

Given,

Heat energy supplied (Q) = 1300 J

Mass of lead (m) = 0.5 kg

Change in temperature (△t) = (40 – 20)°C = 20° C

Specific heat capacity (c) = ?

From relation,

$Q = m \times c \times △t \\[0.5em]$

Substituting the values in the formula above we get,

$1300 = 0.5 \times c \times 20 \\[0.5em] \Rightarrow c = \dfrac{1300}{0.5 \times 20} \\[0.5em] \Rightarrow c = 130 \text{ J kg}^{-1} \text{K}^{-1}$

Hence, specific heat capacity of lead = 130 J kg^-1 K^-1

Given,

Mass of water (m) = 5 kg

Heat absorbed (Q) = 420 kJ = $420\times10^3$ J

Specific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹

Let, increase in temperature be △t.

From relation,

$Q = m c △t$

On rearranging terms,

$△t = \dfrac{Q}{mc}$

Substituting the values in the formula above we get,

$△t = \dfrac{420\times10^3}{5\times4200}=\dfrac{100}{5}=20 K= 20\ ^oC$

Temperature of water will increase by 20 °C.

Given,

Power of heater (P) = 500 W

mass of material (m) = 50 kg

Specific heat capacity of material (c) = 960 J kg^-1 K^-1

Change in temperature △t = (38 – 18)°C = 20° C = 20 K

From relation,

$Q = m c △t$

Substituting the values in the formula above we get,

$Q = 50 \times 960 \times 20 \\[0.5em] \Rightarrow Q = 960,000$

Now,

$Q = \text {Power} \times \text{time}$

Substituting the values in the formula above we get

$960,000 = 500 \times \text{time} \\[0.5em] \Rightarrow \text{time} = \dfrac{960,000}{500} \\[0.5em] \Rightarrow \text{time} = 1920 \text{ s}= 32 \text{ min} \\[0.5em]$

Hence, time taken = 32 min

Given,

Volume of water (V) = 10,000 L

Temperature increased (△t) = 40°C = 40 K

Specific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹

As,

Density of water (d) = 1 Kg/L

$d=\dfrac{m}{V} ⟹ m=dV=1\times10000=10000 \text{ kg}$

Let, Heat absorbed is Q.

From relation,
$Q = m c △t$

Substituting the values in the above formula, $Q = 10,000\times4200\times40=168\times10^7 J$

Heat absorbed by water is 168 x 10⁷ J.

(i) Power of heater (P) = 600 W

Mass of liquid (m) = 4.0 kg

Change in temperature of liquid = (15 – 10)°C = 5° C (or 5 K)

Time taken to raise it's temperature (t) = 100 s

heat capacity = ?

From relation,

$Q = \text {Power} \times \text{time}$

Substituting the values in the formula above we get,

$Q = 600 \times 100 \\[0.5em] \Rightarrow Q = 60000 J \\[0.5em]$

Now,

$C^{'} = \dfrac{Q}{△T} \\[0.5em]$

Substituting the values in the formula above we get,

$C^{'} = \dfrac{60,000}{5} \\[0.5em] C^{'} = 1.2 \times 10^{4} \text{ J} \text{K}^{-1} \\[0.5em]$

Hence, heat capacity = 1.2 x 10 J K^-1

(ii) specific heat capacity {c} = ?

$c = \dfrac{Q}{m \times △T} \\[0.5em]$

Substituting the values in the formula above we get,

$c = \dfrac{60000}{4 \times 5} \\[0.5em] c = \dfrac{60000}{4 \times 5} \\[0.5em] c = 3 \times 10^{3} \text{ J kg}^{-1} \text{K}^{-1}$

Hence, specific heat capacity = 3 x 10³ J Kg^-1 K^-1

Given,

mass (m) = 0.5 kg

Change in temperature = (30 – 5)°C = 25° C = 25 K

Specific heat capacity of squash (c) = 4200 J kg^-1 K^-1

From relation,

$Q = m \times c \times △T \\[0.5em]$

Substituting the values in the formula above we get,

$Q = 0.5 \times 4200 \times 25 \\[0.5em] \Rightarrow Q = 52,500 J$

Let time taken to remove 52500 J of heat be t.

Now it is given that,

30 J of heat is removed in 1 sec

So, 52500 J of heat ⇒ t = ?

$t = \dfrac{1}{30} \times 52,500 \\[0.5em] \Rightarrow t = 1750 \text{ s} \\[0.5em] \Rightarrow t = 29 \text{ min } 10 \text{ sec} \\[0.5em]$

Hence, time taken = 29 min 10 sec

Given,

Mass of metal (m) = 50 g

Fall in temperature of metal = (150 – 20) = 130°C

Rise in temperature of water = (20 - 11) = 9°C

Heat energy given by metal = mc△t
= 50 x c x 130
= 6500 x c     [Equation 1]

Heat energy taken by water = 100 × 4.2 × 9
= 3780     [Equation 2]

Assuming that there is no loss of heat energy,

Heat energy given by metal = Heat energy taken by water.

Equating equations 1 & 2, we get,

$6500 \times c = 3780 \\[0.5em] c = \dfrac{3780}{6500} \\[0.5em] c = 0.582 \text{ J g}^{-1} \text{K}^{-1} \\[0.5em]$

Hence, specific heat capacity of the metal = 0.582 J g^-1 K^-1

Given,

Mass of water = 45 g

Let the final constant temperature reached be t°C

Fall in temperature of water = (50 – t)°C

Mass of copper = 50 g

Rise in temperature of copper = (t - 18)°C

The specific heat capacity of the copper c_c = 0.39 J g^-1 K^-1

The specific heat capacity of water c_w = 4.2 J g^-1 K^-1

Heat energy given by water = mc△t
= 45 x 4.2 x (50 – t)     [Equation 1]

Heat energy taken by copper = 50 x 0.39 x (t - 18)     [Equation 2]

Assuming that there is no loss of heat energy

Heat energy given by water = Heat energy taken by copper

Equating equations 1 & 2, we get,

$45 \times 4.2 \times (50 - t) = 50 \times 0.39 \times (t - 18) \\[0.5em] \Rightarrow 189 \times (50 - t) = 19.5 \times (t - 18) \\[0.5em] \Rightarrow (189 \times 50) - (189 \times t) = (19.5 \times t) - (19.5 \times 18) \\[0.5em] \Rightarrow (9450) - (189 \times t)] = (19.5 \times t) - (351) \\[0.5em] \Rightarrow 9450 + 351 = (19.5 t) + (189 t) \\[0.5em] \Rightarrow 9801 = (208.5t) \\[0.5em] \Rightarrow t = \dfrac{9801}{208.5} \\[0.5em] \Rightarrow t = 47.0072°C \approx 47°C$

Hence, final temperature = 47°C

Mass of hot water = 200 g

Temperature of hot water = 80° C

Mass of cold water = 400 g

Temperature of cold water = 10° C

Let final temperature be t

Fall in temperature of hot water = (80 – t)°C

Rise in temperature of cold water = (t - 10)°C

The specific heat capacity of water c_w = 4200 J kg^-1 K^-1 = 4.2 J g^-1 K^-1

Heat energy given by hot water = mc△t
= 200 x 4.2 x (80 – t)     [Equation 1]

Heat energy taken by cold water = 400 x 4.2 x (t - 10)     [Equation 2]

Assuming that there is no loss of heat energy,

Heat energy given by hot water = Heat energy taken by cold water

Equating equations 1 & 2, we get,

$200 \times 4.2 \times (80 - t) = 400 \times 4.2 \times (t - 10) \\[0.5em] 2 \times (80 - t) = 4\times (t - 10) \\[0.5em] 160 - 2t = 4t - 40 \\[0.5em] 160 + 40 = 4t + 2t \\[0.5em] \\[0.5em] 200 = 6t \\[0.5em] \Rightarrow t = \dfrac{200}{6} \\[0.5em] \Rightarrow t = 33.3°C$

Hence, final temperature = 33.3°C

Given,

mass = 1 kg = 1000 g

Heat energy taken by water (Q) = mc△t
= 1000 x 4.2 x (100 – 25)
= 1000 x 4.2 x 75
= 315000 J

Now,

$Q = \text {Power} \times \text{time}$

Substituting the values in the formula above we get,

$315000 = 1250 \times t \\[0.5em] \Rightarrow t = \dfrac{315000}{1250} \\[0.5em] \Rightarrow t = 252 \text{ sec} = 4 \text{ min } 12 \text{ sec} \\[0.5em]$

Hence, time = 4 min 12 sec

Heat is the internal energy of molecules constituting the body. It flows from a hot body to a cold body when they are kept in contact.

One calorie is the quantity of heat energy required to raise the temperature of 1 g of water from 14.5°C to 15.5°C.

1 calorie (or 1 cal) = 4.186 J or 4.2 J (nearly)

One kilo calorie is the heat energy required to raise the temperature of 1 kg of water from 14.5°C to 15.5°C

Temperature is a parameter which tells the thermal state of a body (i.e., the degree of hotness or coldness of body). It determines the direction of flow of heat when two bodies at different temperatures are placed in contact.

The S.I. unit of temperature is Kelvin (K).

When a hot body is mixed (or is kept in contact) with a cold body, heat energy passes from the hot body to the cold body, till both the bodies attain the same temperature. If no heat is lost to the surrounding then, heat lost by the hot body is equal to the heat gained by the cold body. This is known as the principle of calorimetry.

The term heat capacity of a body is the amount of heat energy required to raise it's temperature by 1° C (or 1 K).

The S.I. unit of heat capacity is joule per kelvin (J K^-1)

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through 1°C (or 1 K). i.e.,

$\text{Specific heat capacity c} = \dfrac{\text{Amt of heat energy supplied}}{\text{mass} \times \text{rise in temperature}}$

The S.I. unit of specific heat capacity is joule per kilogram per kelvin (J kg^-1 K^-1).

(i) "The heat capacity of a body is 50 J K^-1" means 50 J of heat energy is required to raise the temperature of that body by 1 K.

(ii) "The specific heat capacity of copper is 0.4 J g^-1 K^-1" means that the heat energy required to raise the temperature of 1 g of copper by 1 K is 0.4 J.

B is a good conductor of heat energy. For the same heat energy and same mass, the rise in temperature of B will be more hence, B is a good conductor of heat.

The factors on which the heat energy liberated by a body on cooling depends are:

1. Mass of the body
2. Temperature of the body.

Three factors on which the heat energy absorbed by a body depends are:

1. Mass of the body — Heat energy absorbed by a body is directly proportional to the mass of the body i.e., Q ∝ m.
2. Increase in temperature of the body — Heat energy absorbed is directly proportional to the rise in temperature i.e., Q ∝ △t.
3. The material of the body — Heat energy absorbed by a body depends on the substance of the object which is expressed in terms of it's specific heat capacity c i.e., Q ∝ c.

Let,

Specific heat capacity of block P = C_p

Specific heat capacity of block Q = C_Q

From relation,

$c = \dfrac{Q}{m \times △t} \\[0.5em]$

where, c = specific heat capacity

m = mass

Q = heat energy

△t = change in temperature

Now,

$\dfrac{C_p}{C_Q} = \dfrac{\dfrac{Q}{2m × △t}}{\dfrac{Q}{m × △t}} \\[0.5em] \dfrac{C_p}{C_Q} = \dfrac{m × △t}{2m × △t} \\[0.5em] \Rightarrow \dfrac{C_p}{C_Q} = \dfrac{1}{2} \\[0.5em]$

Hence, the required ratio is 1 : 2

Heat energy lost by the hot body = Heat energy gained by the cold body. This is called the principle of method of mixture.

The other name given to it is the principle of calorimetry.

This principle is based on the law of conservation of energy.

On a cold winter night, if the atmospheric temperature falls below 0° C, water in the fine capillaries of plants will freeze, so the veins will burst due to the increase in volume of water on freezing. As a result, plants will die and the crop will get destroyed. In order to save crop on such cold nights, farmers fill their fields with water because water has a high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0° C.

Hot water bottles are used for fomentation because water does not cool quickly due to it's high specific heat capacity, so a hot water bottle provides more heat energy for fomentation over a longer period. Hence, water is used in hot water bottles for fomentation.

Water is an effective coolant. By allowing water to flow in pipes around the heated parts of a machine, heat energy from such parts is removed. Water in pipes can extract more heat from the surroundings without much rise in it's temperature because of it's high specific heat capacity. This is why radiators in car and generator use water for cooling.

The base of a cooking pan is made thick and heavy because it's heat capacity becomes large due to which it gets heated slowly and it imparts sufficient heat energy at a slow rate to the food for it's proper cooking and after cooking it keeps the food warm for a long time.

The S.I. unit of heat is joule (J).

The equation which relates the heat capacity of a body to the specific heat capacity is —
$C=m\times c$

A liquid which has the highest specific heat capacity is water.

The specific heat capacity of water is 4200 J kg^-1 K^-1.

The expression for the heat energy Q is given by:

Q = m x c x △t joules,

where, c is specific heat capacity

m is mass of substance

△t is change in temperature

As the substance having low specific heat capacity will show a rapid and high rise in temperature and since liquid A shows a greater rise in temperature as compared to B when same amount of heat is supplied to both. Hence, heat capacity of liquid A is less than that of B.

(i) Used as a coolant — Radiators in car and generator use water for cooling. Water in car radiator can absorb more heat without much rise in temperature due to it's high specific heat capacity. Hence, in acts as an effective coolant.

(ii) Used as heat reservoir — In cold countries, water is used as heat reservoir for wine and juice bottles to avoid their freezing. The reason is that water due to it's high specific heat capacity can impart a large amount of heat before reaching to it's freezing point. Hence, bottles kept in water remain warm and do not freeze when there is considerable fall in temperature.

(i) Coolant in car radiators — Liquid X is used as a coolant in car radiators because liquid X has specific heat capacity higher than the liquid Y, hence it will absorb more heat energy without much change in temperature.

(ii) Heat reservoir to keep juice bottles without freezing — The liquid needs to give out large amount of heat before reaching freezing temperatures and as liquid X has specific heat capacity higher than the liquid Y, hence liquid X will be used.

(a) The process of change from one state to another at a constant temperature is called the change of phase.

(b) No, there is no change in temperature during the change of phase.

(c) Yes, the substance absorb or liberates heat during the change of phase. The heat is absorbed during melting and it is liberated during freezing at a constant temperature.

(d) The name given to the energy absorbed during a phase change is latent heat.

(a) The process is known as melting.

(b) The heat absorbed by the substance is called the latent heat of melting.

(c) The average kinetic energy of the molecules does not change as there is no change in temperature.

(a) The different parts represent the following:

• Part AB — It shows the rise in temperature of solid from 0°C to t₁°C.
• Part BC — It shows melting at temperature t₁°C.
• Part CD — It shows the rise in temperature of liquid from t₁°C to t₃°C.
• Part DE — It shows boiling at temperature t₃°C

(b) The melting point of the substance is t₁° C.

(c) The boiling point of the substance is t₃° C.

Temperature-time graph representing the cooling of naphthalene from 90°C to room temperature of 25°C is shown below:

Temperature-time graph representing the change of phases when ice is heated till steam is formed at 100°C is shown below:

The change from liquid to gas (or vapour) phase on absorption of heat at a constant temperature is called boiling or vaporisation.

The particular temperature at which vaporisation occurs is known as the boiling point of liquid.

Volume of water increases when it boils at 100° C. 1 cm³ of water at 100 ° C becomes 1760 cm³ of steam at 100 ° C.

(a) The surroundings become pleasantly warm when water in a lake starts freezing in cold countries because the specific latent heat of fusion of ice is very high, hence large quantity of heat is released when the water in the lake freezes. Therefore, the temperature of the surrounding becomes pleasantly warm.

(b) During the change in phase of the substance at a constant temperature on heating, the heat supplied is utilized in increasing the separation against the attractive forces between the molecules. This increases the potential energy of the molecules.
The average kinetic energy of the molecules does not change. Hence, the temperature of the substance remains constant. The heat energy supplied during melting is utilised only in increasing the potential energy of the molecules and is called the latent heat of melting.

absorbed, rejected

Reason — Heat energy is absorbed during melting, as the solid absorbs energy to change into a liquid while maintaining a constant temperature.

Heat energy is rejected during freezing, as the liquid releases energy to change into a solid while maintaining a constant temperature.

decrease

Reason — The melting point of ice decreases when pressure increases because when pressure is increased, volume is decreased and the volume of water is less than ice. So it will be easier to change the state from solid to liquid, and therefore, the melting point decreases.

less than 100°C

Reason — At high altitudes, such as hills and mountains, the atmospheric pressure is low (less than one atmospheric pressure), therefore at these places, water boils at temperature lower than 100° C and so it does not provide the required heat energy to it's contents for cooking. Thus, cooking there becomes very difficult and it takes a much longer time.

higher than 100°C

Reason — The boiling point of water increases by addition of salt to it. If common salt is added to water, it boils at a temperature higher than 100° C.

80 Cal g^-1

Reason — The specific latent heat of fusion of water is 80 Cal g^-1

increasing the potential energy of molecules

Reason — When a substance is melting, it transitions from a solid phase to a liquid phase. During this phase transition, the heat energy supplied is used to overcome the forces holding the molecules together in the solid phase. This process increases the potential energy of the molecules, allowing them to break free from their fixed positions in the crystal lattice. However, the kinetic energy of the molecules remains relatively constant during the phase transition, as the temperature remains constant until the phase change is complete. Therefore, the primary use of heat energy during melting is to increase the potential energy of the molecules

Given,

Mass (m) = 20 g

Heat energy absorbed (Q) = 10,920 J

Specific heat capacity of water = 4200 J kg^-1 K^-1 = 4.2 J g^-1 K^-1

Specific latent heat of fusion of ice (L) = ?

(i) Heat energy required to melt the ice at 0° C to water at 0° C (Q₁) = m x L

Substituting the values in the formula we get,

$Q_1 = 20 \times L$

(ii) Heat energy required to raise temperature from 0° C to 50° C = m x c x rise in temperature

Substituting the values in the formula we get,

$Q_2 = 20 \times 4.2 \times 50 \\[0.5em] = 4200 \text{ J}$

From relation,

$Q = Q_1 + Q_2 \\[0.5em] 10,920 = (20 × \text{ L}) + 4200 \\[0.5em] 10,920 - 4200 = 20 × L\\[0.5em] 6720 = 20 × L\\[0.5em] \Rightarrow L = \dfrac{6720}{20} \\[0.5em] \Rightarrow L = 336 \text{ J g}^{-1} \\[0.5em]$

Hence, Specific latent heat of fusion of ice = 336 J g^-1

Given,

Mass (m) = 500 g

Specific heat capacity of water (c) = 4.2 J g^-1 K^-1

Specific latent heat of fusion of ice (L) = 336 J g^-1

(i) Heat energy released when water lowers it's temperature from 80° C to 0° C

= m x c x change in temperature

Substituting the values in the formula we get,

Q₁ = 500 x 4.2 x (80 - 0)

= 500 x 4.2 x 80

= 168 x 10³ J

(ii) Heat energy released when water at 0° C changes into ice at 0° C = m x L

Substituting the values in the formula we get,

Q₂ = 500 x 336

= 168 x 10³ J

From relation,

Q = Q₁ + Q₂

= 168 x 10³ J + 168 x 10³ J

= 336 x 10³ J

Total heat released = 336 x 10³ J

(a) Given,

Mass (m) = 150 g

Heat energy given out (Q) = 75,000 J

Specific latent heat of the metal = ?

From relation Q = m x L

Substituting the values in the formula we get,

$75000 = 150 \times L \\[0.5em] \Rightarrow L = \dfrac{75000}{150} \\[0.5em] \Rightarrow L = 500 \text{ J g}^{-1} \\[0.5em]$

Hence, the specific latent heat of the metal = 500 J g^-1

(b) Specific heat capacity of metal is 200 J kg^-1 K^-1

Change in temperature
= 800 – (-50) = 800 + 50 = 850° C = 850 K

From relation,

Q = m x c x change in temperature

Substituting the values we get,

$Q = 0.15 \times 200 \times 850 \\[0.5em] Q = 25,500 J \\[0.5em]$

Hence, 25,500 J of heat energy will be given out

Given,

m = 150 g = 0.15 kg

P = 100 W

t = 4 min = 240 s

specific latent heat of fusion of the metal = ?

Heat supplied = P x t

Substituting the values we get,

$\text{Heat supplied} = 100 \times 240 \\[0.5em] \text{Heat supplied} = 24000 J$

We know,

Q = m x L

Substituting the values we get,

$24000 = 0.15 \times \text{L} \\[0.5em] \Rightarrow \text{L} = \dfrac{24000}{0.15} \\[0.5em] \Rightarrow \text{L} = 160,000 \\[0.5em] \Rightarrow \text{L} = 1.6 \times 10^{5} \text{ J kg}^{-1}$

Hence, specific latent heat of fusion of the metal = 1.6 x 10⁵ J kg^-1

Given,

mass (m) = 100 g

time (t) = 73.5 min

specific heat capacity of water = 4.2 J g^-1 K^-1

specific latent heat of ice = 336 J g^-1

specific heat capacity of ice = 2.1 J g^-1 K^-1

Heat energy released by water in fall of it's temperature from 20° to 0° C (Q₁)
= mass x specific heat capacity x fall in temperature
= 100 × 4.2 x (20 - 0)
= 100 × 4.2 x 20
= 8400 J

Hence, Q₁ = 8400 J

Heat energy released by water when it converts into ice at 0° C (Q₂) = m x L_ice
= 100 × 336
= 33600 J

Heat energy released when ice cools from 0° C to -10° C (Q₃) = m x c x change in temperature
= 100 × 2.1 x [0 - (-10)]
= 100 x 2.1 x 10
= 2100 J

Hence,

Total heat energy = Q₁ + Q₂ + Q₃
= 8400 + 33600 + 2100
= 44100 J

Time taken = 73.5 min = 4410 s

Average rate of heat extraction (P)

$P = \dfrac{E}{t}$

Substituting the values in the formula we get,

$P = \dfrac{44100}{4410} \\[0.5em] P = 10 W$

Hence, average rate of heat extraction = 10 W

Given,

Mass of ice (m₁) = 17 g

Mass of water (m₂) = 40 g

Change in temperature = 34 – 0 = 34° C = 34 K

Specific heat capacity of water (c) = 4.2 J g^-1 K^-1

specific latent heat of ice = ?

Assuming that no heat energy is lost,

heat energy required by ice to melt = heat energy given by water

So,

m₁ x L = m₂ x c x change in temperature

Substituting the values in the relation above we get,

$17 \times L = 40 \times 4.2 \times 34 \\[0.5em] 17 \times L = 5712 \\[0.5em] \Rightarrow L = \dfrac{5712}{17} \\[0.5em] \Rightarrow L = 336 \text{ J g}^{-1} \\[0.5em]$

Hence, specific latent heat of ice = 336 J g^-1

Assumption — There is no loss of energy.

Given,

m_w = 170 g = 0.17 kg

specific heat capacity of water = 4200 J kg^-1 K^-1

specific latent heat of ice = 336000 J kg^-1

m_i = ?

Heat energy given out by water in lowering it's temperature from 50° C to 5° C
= m x c x change in temperature
= 0.17 x 4200 x (50 - 5)
= 0.17 x 4200 x 45
= 32,130

Heat energy taken by m kg ice to melt into water at 0° C
= m_i x L
= m_i x 336000

Heat energy taken by water at 0° C to raise it's temperature to 5° C
= m_i x c x change in temperature
= m_i x 4200 x (5 - 0)
= m_i x 4200 x 5
= m_i x 21000

heat energy released = heat energy taken

Substituting the values we get,

$32130 = (m_i \times 336000 ) + (m_i \times 21000 ) \\[0.5em] 32130 = m_i \times 357000 \\[0.5em] \Rightarrow m_i = \dfrac{32130}{357000} \\[0.5em] \Rightarrow m_i = 0.09 \text{ Kg} = 90 \text{ g}$

Hence, the mass of ice added = 90 g

Given,

mass of ice = 10 g

mass of water = 10 g

Specific heat capacity of ice = 2.1 J g^-1 K^-1

specific latent heat of ice = 336 J g^-1,

specific heat capacity of water = 4.2 J g^-1 K^-1

Let the final temperature be t° C.

Heat energy taken by ice at – 10° C to raise it's temperature to 0° C (Q₁)
= m x c x change in temperature
= 10 x 2.1 x [0 - (-10)]
= 10 x 2.1 x 10
= 210 J

Hence, Q₁ = 210 J

Heat energy taken by ice at 0° C to convert into water at 0° C (Q₂)
= m x L_ice
= 10 × 336
= 3360 J

Hence, Q₂ = 3360 J

Heat energy taken by water at 0° C to raise it's temperature to t° C (Q₃) = m x c x change in temperature
= 10 × 4.2 × (t – 0)
= 10 × 4.2 × t
= 42 t

Hence, Q₃ = 42 t

Heat energy released by water at 10° C to lower it's temperature to t° C (Q₄) = m x c_water x change in temperature
= 10 × 4.2 × (10 – t)
= 42 × (10 – t) = 420 – 42t

Hence, Q₄ = 420 – 42t

If there is no loss of heat,

Heat energy gained = Heat energy lost

$210 + 3360 + 42 t = 420 - 42t \\[0.5em] 3570 - 420 = - 42 t - 42 t \\[0.5em] 3150 = - 84 t \\[0.5em] t = - \dfrac{3150}{84} \\[0.5em] t = - 37.5° C$

This cannot be true because water cannot exist at -37.5° C.

Hence, we can say that the whole of the ice did not melt.

Let amount of ice which melts = m g

Final temperature of the mixture = 0° C

Heat energy gained by ice at -10° C to raise it's temperature to 0° C

= m x c_ice x change in temperature
= 10 × 2.1 x [0 - (-10)]
= 10 × 2.1 x 10
= 210 J

Heat energy gained by m gm of ice at 0° C to change into water at 0° C
= m × L_ice
= m x 336
= 336m J

Heat energy released by 10 g of water at 10° C to lower it's temperature to 0° C
= m x c_water x change in temperature
= 10 × 4.2 × (10 – 0)
= 10 × 4.2 × 10 = 420 J

If there is no loss of heat,

Heat energy gained = Heat energy lost

$210 + 336 m = 420 \\[0.5em] \Rightarrow 336 m = 420 - 210 \\[0.5em] \Rightarrow 336 m = 210 \\[0.5em] \Rightarrow m = \dfrac{210}{336} \\[0.5em] \Rightarrow m = 0.625 g$

Hence, 0.625 g of ice will melt and temperature will remain at 0° C

Given,

m_i = 40 g

m_w = 200 g

Specific heat capacity of water = 4200 J kg^-1 K^-1

specific latent heat of fusion of ice = 336 x 10³ J kg^-1

final temperature of water = ?

Let final temperature = t

Heat energy given out by water when it cools from 50° C to t° C
= m x c x change in temperature
= 200 × 4.2 × (50 – t)
= 42000 – 840t

Heat energy taken by ice when it converts from ice into water at 0° C
= m x L
= 40 × 336 J
= 13440 J

Heat energy taken by water when it raises it's temperature from 0° to t° C
= m x c x change in temperature
= 40 × 4.2 × (t – 0)
= 40 × 4.2 × t
= 168t

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$13440 + 168t = 42000 – 840t \\[0.5em] 168t + 840t = 42000 – 13440 \\[0.5em] 1008t = 28560 \\[0.5em] t = \dfrac{28560}{1008} \\[0.5em] t = 28.330° C$

Hence, final temperature = 28.330° C

Given,

mass of water m_w = 150 g

mass of calorimeter m_c = 50 g

Specific heat capacity of calorimeter = 0.4 J g^-1 °C^-1,

Specific heat capacity of water = 4.2 J^-1 °C^-1,

Latent heat capacity of ice = 330 J g^-1

mass of ice m_i = ?

Heat energy imparted by calorimeter and water contained in it in cooling from 32° C to 5° C is used in melting ice and then raising the temperature of melted ice from 0°C to 5°C.

Heat energy imparted by water (Q₁)
= m x c x change in temperature
= 150 x 4.2 x (32 - 5)
= 150 x 4.2 x 27
= 17,010 J

Heat energy imparted by calorimeter (Q₂)
= m x c x change in temperature
= 50 x 0.4 x (32 - 5)
= 50 x 0.4 x 27
= 540 J

Heat energy taken by ice to melt (Q₃)
= m_i x L
= m_i x 330

Heat energy gained by water from melted ice to reach from 0°C to 5°C (Q₄)
= m_i x c x change in temperature
= m_i x 4.2 x (5 - 0)
= m_i x 4.2 x 5
= m_i x 21

From the principle of calorimetry, if the system is fully insulated then,

Heat gained by cold body = Heat lost by hot body.

Q₁ + Q₂ = Q₃ + Q₄

Substituting the values we get,

$17010 + 540 = (m_i \times 330) + (m_i \times 21) \\[0.5em] 17550 = m_i \times 351 \\[0.5em] \Rightarrow m_i = \dfrac{17550 }{351} \\[0.5em] \Rightarrow m_i = 50 g$

Hence, mass of ice = 50 g

Given,

m_copper = 50 g

m_water = 250 g

Final temperature = 5° C.

Let mass of ice required be m_i.

Heat energy gained by ice at 0° C to convert into water at 0° C
= m_i × L
= m_i x 336 J

Heat energy gained by (m) g of water at 0° C to rise it's temperature to 5° C
= m x c x change in temperature
= m_i x c x (5 - 0) = m_i × 4.2 × 5
= 21 x m_i

Heat energy lost by water at 30° C in cooling to 5° C
= m x c x change in temperature
= 250 × 4.2 × (30 - 5)
= 250 × 4.2 × 25
= 26250 J

Heat energy lost by vessel at 30° C to cool down to 5° C =
= m x c x change in temperature
= 50 × 0.4 × (30 - 5)
= 50 × 0.4 × 25
= 500 J

If there is no loss of heat,

Heat energy gained = Heat energy lost

Substituting the values in the relation above we get,

$(336 \times m_i ) + (21 \times m_i) = 26250 + 500 \\[0.5em] 357 \times m_i = 26750 \\[0.5em] \Rightarrow m_i = \dfrac{26750}{357} \\[0.5em] \Rightarrow m_i = 74.93 g$

Hence, required mass of ice = 74.93 g

Given,

m_i = 2 kg

Specific heat capacity of water = 4200 J kg^-1 K^-1,

Specific latent heat of ice = 336 × 10³ J Kg^-1.

m_water = ?

Since the whole block does not melt and only 2 kg of it melts, so final temperature would be 0° C.

Heat energy taken by ice at 0° C to convert into water at 0° C
= m x L
= 2 × 336000
= 672000 J

Initial temperature of water = 100° C

Final temperature of water = 0° C

Heat energy lost by (m_w) kg at 100° C to reach temperature 0° C
= m x c x change in temperature
= m × 4200 × (100 - 0)
= m × 4200 × 100
= 420000 m J

If there is no loss of energy,

heat energy gained = heat energy lost

Substituting the values in the relation above we get,

$672000 = m × 420000 \\[0.5em] m = \dfrac{672000}{420000} \\[0.5em] m = 1.6 kg \\[0.5em]$

Hence, mass of water = 1.6 kg

Given,

m = 100 g

Specific heat capacity of ice = 2.1 J g^-1 K^-1

specific heat capacity of water = 4.2 J g^-1 K^-1

specific latent heat of ice = 336 J g^-1

total amount of heat energy required = ?

Heat energy taken by ice to raise it's temperature from at –10° C to 0° C
= m x c x change in temperature
= 100 × 2.1 × [0 - (-10)]
= 100 x 2.1 x 10
= 2100 J

Heat energy taken by ice at 0° C to convert into water at 0° C
= m x L
= 100 × 336
= 33600 J

Heat energy taken by water to raise the temperature from 0° C to 100° C
= m x c x change in temperature
= 100 × 4.2 × (100 - 0)
= 100 × 4.2 × 100
= 42000 J

Total heat energy gained is
= 2100 + 33600 + 42000
= 77700 J
= 7.77 × 10⁴ J

Hence, total amount of heat energy required = 7.77 × 10⁴ J

Given,

m = 1 kg

heat energy required = 7,77,000 J

Specific heat capacity of ice = 2100 J kg^-1 K^-1

Specific heat capacity of water = 4200 J kg^-1 K^-1

specific latent heat of ice (L) = ?

Heat energy taken by ice to raise temperature from – 10° C to to 0° C
= m x c x change in temperature
= 1 × 2100 × [0 - (-10)]
= 21000 J

Heat energy gained by ice at 0° C to convert into water at 0° C
= m x L
= 1 x L
= L

Heat energy taken by water to raise it's temperature from 0° C to 100° C
= m x c x change in temperature
= 1 x 4200 x (100 - 0)
= 1 × 4200 × 100
= 4,20,000 J

Total heat energy gained
= 21,000 + L + 4,20,000 = 4,41,000 + L

As,

4,41,000 + L = 7,77,000
L = 7,77,000 – 4,41,000
L = 3,36,000 J kg^-1

Hence, specific latent heat of ice = 3,36,000 J kg^-1

Given,

m_i = 200 g

m_w = 200 g

Time for ice to melt (t₁) = 1 min = 60 s

Change in temperature of water (Δt) = 20° C

specific latent heat of ice = 336 J g^-1

Rate of heat exchange is constant.

Therefore, power required for converting ice to water = power required to increase the temperature of water.

$P_i = P_w \\[0.5em] \dfrac{E_i}{t_1} = \dfrac{E_w}{t_2} \\[0.5em] \dfrac{ m_i \times L}{t_1} = \dfrac{ m_w \times c_w \times Δt }{t_2} \\[0.5em] \dfrac{ 200 \times 336}{60} = \dfrac{ 200 \times 4.2 \times 20 }{t_2} \\[0.5em] \Rightarrow t_2 = \dfrac{5040}{336} \\[0.5em] \Rightarrow t_2 = 15 s$

Hence, time taken = 15 s

Given,

Mass of sweat lost (m) = 1 kg

Latent heat of vaporization (L) = 2268 × 10³ J kg⁻¹

Specific heat capacity of water (c) = 4.2 × 10³ J kg⁻¹ K⁻¹

Now,

Energy lost by body through evaporation (Q) = mL = $1\times2268\times10^3=2268\times10^3\ J$

And,

To raise 1 kg of water by 1 °C, energy required (q) = mcΔt

On putting values,

$q=1\times4.2\times10^3\times1= 4.2\times10^3\ J$

So, evaporation of 1 kg of sweat causes heat loss equivalent to increasing the temperature of 1 kg of water by :

Equivalent temperature rise = $\dfrac{Q}{q}=\dfrac{2268\times 10^3}{4.2\times 10^3}=540\ ^oC$

The cooling effect of evaporation of 1 kg of sweat is equivalent to cooling 1 kg of water by 540°C. This shows that evaporation causes a much greater cooling effect compared to simple heating or cooling based on specific heat capacity.

(i) When a substance on heating undergoes a rise in it's temperature then average kinetic energy of molecules increases.

(ii) When a substance on heating undergoes a change in it's phase without change in it's temperature then average potential energy of molecules increases.

(a) The average kinetic energy of molecules does not change during it's change in phase at a constant temperature, on heating.

(b) Average potential energy of molecules increases during it's change in phase at a constant temperature, on heating.

Explanation: During the change in phase of the substance at a constant temperature on heating, the heat supplied is utilized in increasing the separation against the attractive forces between the molecules. This increases the potential energy of the molecules.
As the temperature of the substance remains constant, the average kinetic energy of the molecules does not change. The heat energy supplied during melting is utilised only in increasing the potential energy of the molecules and is called the latent heat of melting.

The melting point of a substance decreases by the presence of impurities in it. The melting point of ice decreases from 0 ° C to -22 ° C on mixing salt to it in proper proportion. This fact is utilized in making the freezing mixture by adding salt to ice. The freezing mixture is used in preparing 'kulfis'.

The melting point of the substances which contract on melting (like ice) decreases by the increase in pressure. For example, the melting point of ice decreases by 0.0072°C for every one atmosphere rise in pressure.

The boiling point of water increases by addition of salt to it. If common salt is added to water, it boils at a temperature higher than 100° C.

The boiling point of a liquid increases with the increase in pressure.

The boiling point of liquid increases with the increase in pressure and decreases with the decrease in pressure. The boiling point of pure water at one atmospheric pressure is 100° C.

In a pressure cooker, steam is not allowed to escape out. The vapour pressure on water inside the cooker becomes nearly 1.75 times the atmospheric pressure, so water boils in it at about 120° C to 125° C due to increased pressure.

At high altitudes, such as hills and mountains, the atmospheric pressure is low (less than one atmospheric pressure), therefore at these places, water boils at temperature lower than 100° C and so it does not provide the required heat energy to it's contents for cooking. Thus, cooking there becomes very difficult and it takes a much longer time.

Heat energy absorbed or liberated in change of phase that is not externally manifested by any rise or fall in temperature is called the latent heat.

The specific latent heat of fusion of ice is the heat energy required to melt unit mass of ice at 0° C to water at 0° C without any change in temperature.

S.I. unit of latent heat of fusion is J kg^-1

'The specific latent heat of fusion of ice is 336 J g^-1' means 1 g of ice at 0° C absorbs 336 J of heat energy to convert into water at 0° C.

1 g of water at 0° C has more heat because 1 g of water at 0° C liberates 80 cal heat to form 1 g of ice at 0° C.

(a) 1 g ice at 0° C requires more heat to raise it's temperature to 10° C.

(b) 1 g ice at 0° C requires more heat to raise it's temperature to 10° C because 1 g ice at 0° C first absorbs 336 J heat to convert into 1 g water at 0° C and then the water absorbs heat to raise it's temperature from 0° to 10° C.

Ice cream absorbs heat energy as well as the latent heat while water absorbs only heat energy. Therefore, ice cream absorbs more amount of energy from the mouth as compared to water. Hence, ice cream appears colder to the mouth than water at 0° C.

Ice cubes at 0°C will cool the soft drink bottles more quickly than iced-water at 0°C.

1 g of ice at 0° C takes 336 J of heat energy from the drink to melt into water at 0° C. Thus, the drink liberates an additional 336 J of heat energy to 1 g ice at 0° C than to 1 g ice-cold water at 0° C. Therefore, cooling produced by 1 g ice at 0° C is much more than that by 1 g water at 0° C.

It is generally cold after a hail storm than during and before it because after the hail storm, ice absorbs the heat energy required for it's melting from the surroundings, so the temperature of the surrounding falls further down and we feel more cold.

The temperature of surroundings starts falling when ice in a frozen lake starts melting because quite a large amount of heat energy is required for melting the frozen lake which is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it becomes very cold.

Water in lakes and ponds do not freeze at once in cold countries because the specific latent heat of fusion of ice is sufficiently high (= 336 J g^-1). The water in lakes and ponds will have to liberate a large quantity of heat to the surrounding before freezing. The layer of ice formed over the water surface, being a poor conductor of heat, will also prevent the loss of heat from the water of lake, hence the water does not freeze all at once.

The water boils at about 120° C to 125° C in a pressure cooker.

1. When ice melts, it's volume decreases.
2. Decrease in pressure over ice increases its melting point.
3. Increase in pressure increases the boiling point of water.
4. A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure
5. The boiling point of water is defined as the constant temperature at which water changes to steam
6. Water can be made to boil at 115°C by increasing pressure over it's surface.

The approximate value of specific latent heat of ice is 336 x 10³ J kg^-1

Latent heat is absorbed by ice when 1 g ice at 0° C melts to form 1 g water at 0° C.