CBSE Class 12 Physics Calorimetry Theory and Examples

CBSE Class 12 Physics Calorimetry Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Calorimetry is the branch of heat that deals with the measurement of heat. This Class 12 Physics chapter notes page covers specific heat of solids, liquids and gases, molar and gram specific heats, thermal capacity, water equivalent of a body, latent heat of fusion and vaporisation, and the specific heat of a gaseous mixture, with solved CBSE-style examples on each topic.

Calorimetry

Calorimetry is that branch of heat which deals with its measurement. The usual units of heat are calorie or kilocalorie.

1. SPECIFIC HEAT

This is also called 'Heat Inertia' of a substance.

1.1 Specific Heat of Solid and Liquid (s)

If a solid or a liquid is heated till the change of state, there is no change in its volume, so the work done is zero. Hence there is only one specific heat of solids and liquids, and it is constant.
1. Definition: The amount of heat needed for a unit increase in the temperature of unit mass of a solid or liquid is called its specific heat.
Unit: kilocalorie/kg-°C or calorie/g-°C
1 kcal/kg-°C = 1 cal/g-°C
2. If the mass of a body is m and its specific heat is s, the amount of heat needed to increase its temperature by dT is Q = msdT.
3. Specific heat of water = 1 kcal/kg-°C = 1 cal/g-°C = 4.18 × 10³ J/kg-°C.
4. Kelvin can also be used instead of °C, since the size of both units is the same.

1.2 Specific Heat of Gas (C)

1. There are many processes possible to give heat to a gas. A specific heat can be associated with each such process, and it depends on the nature of the process.
2. The number of possible specific heats for a gas is infinite, and their values can vary from zero (0) to infinity (∞).
3. Generally two types of specific heat are mentioned for a gas:
(a) specific heat at constant volume (Cv)
(b) specific heat at constant pressure (Cp)
4. These specific heats can be molar or gram, depending on the amount of gas considered to define it.
5. The molar heat capacity of a gas is defined as the heat given per mole of the gas per unit rise in temperature.
6. The molar heat capacity at constant volume, denoted by Cv, is Cv = \( \frac{\Delta Q}{n \Delta T} \).
7. If, at absolute temperature T, the total energy of a gas is E and the degree of freedom of the gas is f, then E = \( \frac{fRT}{2} \), Cv = \( \frac{\Delta E}{\Delta T} \) and Cv = \( \frac{Rf}{2} \).
8. The amount of heat needed to increase the temperature of n moles of gas by dT at constant volume is dθ = ncv dT.
9. The amount of heat needed to increase the temperature of 1 g of gas by 1°C at constant volume is called the gram specific heat at constant volume.
10. (Cv)gram = \( \frac{(C_v)_{molar}}{M} \), or (Cv)molar = M(Cv)gram, where M is the molecular weight of the gas.
11. The amount of heat needed to increase the temperature of m gram of gas by dT at constant volume is dQ = m(Cv)gram dT = \( \frac{m}{M} \)(Cv)molar dT.
12. The amount of heat needed to increase the temperature of 1 mole of gas by 1°C at constant pressure is called the molar specific heat at constant pressure.
13. Cp = Cv + R, for one mole, where R is the universal gas constant.
14. The amount of heat needed to increase the temperature of n moles of gas by dT at constant pressure is dQ = nCp dT.
Note: If a gas is heated at constant pressure, Cv can be replaced by Cp in the above discussion. Hence (Cp)gram = \( \frac{C_p}{M} \), or Cp = M(Cp)gram.
15. Specific heat for other processes is S = \( \frac{Q}{m \Delta T} \).
(a) Adiabatic: S = 0, as Q = 0 but ΔT has some value.
(b) Isothermal: S = ∞, as ΔT = 0 but Q has some value.
16. Specific heat for any process is given by C = Cv + \( \frac{p\,dV}{n\,dT} \) = Cp - R + \( \frac{p\,dV}{n\,dT} \), for n moles of gas.

Example:
(1) If the process is isochoric, then dV = 0, so C = Cv.
(2) If the process is isobaric, then pV = nRT, so \( \frac{p\,dV}{dT} \) = nR. Hence C = Cv + R = Cp, for one mole.

2. THERMAL CAPACITY

1. The amount of heat needed to increase the temperature of a substance (of any amount) by 1°C is called the thermal capacity of that substance.
2. Thermal capacity = (mass of body) × (specific heat), so Hc = ms.
3. Unit: calorie/°C or kcal/°C.
4. Heat capacity at a point p on the heat-temperature curve is \( \frac{1}{\tan \theta} \) = cot θ. The thermal capacity is the reciprocal of the slope of the heat-temperature curve, so Hc = ms = \( \frac{Q}{dT} \).
5. Heat capacity in an isothermal process is infinite (∞). For example, during melting and vaporisation.
6. If the heat capacity of a body is Hc, then the heat needed to rise its temperature by dθ is dQ = Hc dθ.

3. WATER EQUIVALENT OF A BODY

1. If Q amount of heat raises the temperature of m gram of a substance by ΔT, and the same heat raises the temperature of w gram of water by the same ΔT, then w is called the water equivalent of the body of mass m.
2. The water equivalent of a body is the same as its heat capacity. The only difference is in the units. For example, if the heat capacity of a body is m calorie/°C, then its water equivalent is m gram.
3. Physical meaning: the same amount of heat has to be given to a body to raise its temperature by dT as is needed for a quantity of water equal to its water equivalent, over the same temperature range.

4. LATENT HEAT

Latent heat of fusion of a substance is the quantity of heat (in kilocalories) required to change 1 kg of its mass from solid to liquid state at its melting point. For ice, the latent heat of fusion is 80 kilocal/kg.
Latent heat of vaporisation of a substance is the quantity of heat required to change 1 kg of its mass from liquid to vapour state at its boiling point. For water, the latent heat of vaporisation is 536 kilocal/kg.

5. GASEOUS MIXTURE

(i) Suppose two non-reactive gases are enclosed in a vessel of volume V. If n1 moles of one gas are mixed with n2 moles of another gas, and N_A is Avogadro's number, then the number of molecules of the first gas is N1 = n1N_A and the number of molecules of the second gas is N2 = n2N_A.
(ii) Total number of moles n = (n1 + n2).
(iii) If M1 is the molecular weight of the first gas and M2 that of the second gas, the molecular weight of the mixture is M = \( \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2} \).
(iv) Specific heat of the mixture at constant volume is Cv = \( \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \).
(v) Specific heat of the mixture at constant pressure is Cp = \( \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} \).
(vi) Ratio of specific heats of the mixture is γ = \( \frac{C_p}{C_v} \) = \( \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} \).
(vii) Pressure exerted by the mixture is P = (n1 + n2) \( \frac{RT}{V} \) = (\( \frac{m_1}{M_1} \) + \( \frac{m_2}{M_2} \)) \( \frac{RT}{V} \), where m1 and m2 are the masses of the two gases.
(viii) If n1 moles of the first gas at temperature T1 are mixed with n2 moles of the other gas at temperature T2, the temperature of the mixture is T = \( \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \).

Examples based on Specific Heat, Thermal Capacity and Latent Heat

Ex.1 The amount of heat necessary to raise the temperature of 0.2 mol of N₂ at constant pressure from 37º to 337º C will be -
(A) 1764 Joule
(B) 764 Joule
(C) 1764 Calorie
(D) 1764 erg

Answer: (A) 1764 Joule

1. Heat at constant pressure is dQ = nCp dT.
2. For a diatomic gas such as N₂, Cp = 7 cal/(mol K), using R = 2 cal/(mol K) and Cp = 7R/2.
3. Temperature change dT = 337 - 37 = 300 K.
4. dQ = 0.2 × 7 × 300 = 420 cal.
5. Converting to joules, dW = 420 × 4.2 = 1764 J.
6. Hence, the correct option is (A).

Teacher's Note:
a) Always convert the temperature difference, not the absolute temperatures, into kelvin.
b) Use 4.2 J per calorie to convert the heat into joules.

Ex.2 (Cp - Cv) will be -
(A) Maximum for oxygen
(B) Maximum for nitrogen
(C) Maximum for carbon di oxide
(D) Same for all gases

Answer: (D) Same for all gases

1. For one mole of gas, Cp - Cv = R.
2. R is the universal gas constant, and its value is the same for every gas.
3. Hence, (Cp - Cv) is the same for all gases, and the correct option is (D).

Teacher's Note:
a) Cp - Cv = R holds only for one mole of an ideal gas.
b) The difference is not the same as the ratio Cp/Cv, which does vary from gas to gas.

Ex.3 The ratio of radii of two copper spheres is 2 : 1 and they are kept at same temperature. The ratio of their heat capacities will be -
(A) 2 : 1
(B) 1 : 1
(C) 8 : 1
(D) 4 : 1

Answer: (C) 8 : 1

1. Heat capacity H = ms, where m = \( \frac{4}{3} \) πr³ρ for a sphere.
2. So H = \( \frac{4}{3} \) πr³ρS.
3. Ratio H1 : H2 = r1³ : r2³ = 2³ : 1³ = 8 : 1.
4. Hence, the correct option is (C).

Teacher's Note:
a) Heat capacity of a solid sphere depends on its volume.
b) Doubling the radius increases the volume eight times.
c) Both spheres are copper, so the density and specific heat are the same and cancel out.

Ex.4 5 g of ice at 0°C is dropped ina beaker containing 20 g of water at 40°C, then
(A) All the ice will not melt into water
(B) All the ice will melt and the resulting temperature of water will be 0°C
(C) All the ice will melt and the resulting temperature of water will be 25°C
(D) All the ice will melt and the resulting temperature of water will be 16°C

Answer: (D) All the ice will melt and the resulting temperature of water will be 16°C

1. Heat needed to melt 5 g of ice = 5 × 80 = 400 cal.
2. Heat released by 20 g of water cooling from 40°C to 0°C = 20 × 1 × 40 = 800 cal.
3. After melting the ice, the remaining heat is 800 - 400 = 400 cal.
4. This heat warms the total water mass of 25 g, so 20 × 1 × (40 - θ) = 5 × 80 + 5 × 1 × (θ - 0).
5. 800 - 20θ = 400 + 5θ, so 25θ = 400 and θ = 16°C.
6. Hence, the correct option is (D).

Teacher's Note:
a) Check first whether the heat from the water is enough to melt all the ice.
b) Melted ice at 0°C adds to the water mass, so use 25 g in the last step.
c) The final temperature lies between 0°C and 40°C, which rules out the other options.

Examples based on Gaseous Mixture

Ex.5 n₁ and n₂ moles of two ideal gases of the thermodynamics constants γ1 and γ2 respectively are mixed. C_P/C_v for the mixture is -
(A) \( \frac{\gamma_1 + \gamma_2}{2} \)
(B) \( \frac{n_1 \gamma_1 + n_2 \gamma_2}{n_1 + n_2} \)
(C) \( \frac{n_1 \gamma_2 + n_2 \gamma_1}{n_1 + n_2} \)
(D) \( \frac{n_1 \gamma_1 (\gamma_2 - 1) + n_2 \gamma_2 (\gamma_1 - 1)}{n_1 (\gamma_2 - 1) + n_2 (\gamma_1 - 1)} \)

Answer: (D) \( \frac{n_1 \gamma_1 (\gamma_2 - 1) + n_2 \gamma_2 (\gamma_1 - 1)}{n_1 (\gamma_2 - 1) + n_2 (\gamma_1 - 1)} \)

1. For ideal gases at the same temperature, the internal energy of the mixture is U = U1 + U2, so nCv = n1Cv1 + n2Cv2.
2. For an ideal gas, Cv = \( \frac{R}{\gamma - 1} \) and Cp = \( \frac{\gamma R}{\gamma - 1} \).
3. Substituting, nCv = n1 \( \frac{R}{\gamma_1 - 1} \) + n2 \( \frac{R}{\gamma_2 - 1} \) and nCp = n1 \( \frac{\gamma_1 R}{\gamma_1 - 1} \) + n2 \( \frac{\gamma_2 R}{\gamma_2 - 1} \).
4. Dividing, γ = \( \frac{n_1 \gamma_1 (\gamma_2 - 1) + n_2 \gamma_2 (\gamma_1 - 1)}{n_1 (\gamma_2 - 1) + n_2 (\gamma_1 - 1)} \).
5. Hence, the correct option is (D).

Teacher's Note:
a) Ratio of specific heats of a mixture is not a simple average of the two γ values.
b) Express Cv and Cp in terms of γ before mixing.
c) Option (B) looks similar, but it averages γ directly, which is wrong.

Ex.6 One mole of monatomic gas (γ = 5/3) is mixed with one mole of a diatomic gas (γ = 7/3), the value of γ for the mixture is
(A) 1.40
(B) 1.50
(C) 1.53
(D) 3.07

Answer: (B) 1.50

1. For the monatomic gas, Cp = \( \frac{5R}{2} \) and Cv = \( \frac{3R}{2} \).
2. For the diatomic gas, Cp = \( \frac{7R}{2} \) and Cv = \( \frac{5R}{2} \), which corresponds to γ = 7/5.
3. γ = \( \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} \) = \( \frac{5/2 + 7/2}{3/2 + 5/2} \) = \( \frac{6}{4} \) = 1.5.
4. Hence, the correct option is (B).

Teacher's Note:
a) The printed value γ = 7/3 for a diatomic gas is a typo for 7/5, the correct value for a diatomic gas.
b) The official key shows (C) 1.53, but its own working gives 1.5, which is option (B).
c) Monatomic gases have γ = 5/3 and diatomic gases have γ = 7/5.

⇒ E =2fRT and Cv =ΔE/ΔT

⇒ CV =Rf/2

(10) (C_v)gram =M(C_v )molar  or (C_v)molar = M(C_v)gram
where M = molecules wt . of gas

(2) The value of water equivalent of a body is same as it's heat capacity. The difference is only in units. e.g If heat capacity of a body is m calorie/ 0C then it's water equivalent will be m gram.