HC Verma Concepts Of Physics PDF Exercise Solutions Of Chapter 25 (Calorimetry)

Solutions of H.C. Verma’s Concepts of Physics chapter 25 (Calorimetry) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 25 (Calorimetry)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 25 (Calorimetry):

EXERCISE

1. An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J/kg-K, 470 J/kg-K and 4200 J/kg-K respectively.

Sol:

Given: Mass of aluminium m_a = 0.5kg; Mass of water m_w = 0.2 kg; Mass of Iron m_i = 0.2 kg; Temperature of aluminium and water = 20°C = 297 K; Temperature of Iron = 100°C = 373 K; Specific heat of aluminium = 910 J/kg-k; Specific heat of Iron = 470 J/kg-k; Specific heat of water = 4200J/kg-k.

Assumption: heat interaction outside of the boundary is zero.

∴ Heat loss by iron block = heat gain by aluminium vessel and water

Or, 0.2 * 470 * (373 – T) = (T – 293) (0.5 * 910 + 0.2 * 4200)

Or, 94 * (373 – T) = (T – 293) (455 + 840)

Or, 373 – T = (T – 293) * 13.8

Or, T = 298 K = 25 ⁰C.

2. A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J/kg-°C.

Sol:

Given: mass of iron, m_iron = 100 g = 0.1 kg; mass of water, m_w = 240 g = 0.24 kg; water equivalent of calorimeter, m_eq = 10 g = 0.01 kg; final temp of mixture, T_f = 60⁰ C; temp of water, T_w = 20⁰ C; Specific heat capacity of iron, C_iron = 470 J/kg-°C; C_w = 4184 J/kg-°C.

Let the furnace temperature be T.

Now, heat loss by the iron = heat gain by the water and calorimeter

Or, m_iron * C_iron * (T – T_f) = (m_w + m_eq) * C_w * (T_f – T_w)

Or, 0.1 * 470 * (T – 60) = (0.24 +0.01) * 4184 * (60 – 20)

Or, T = 950⁰ C.

So, the temperature of the furnace is 950⁰ C. 

3. The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed?

Sol:

Given: T_A = 12⁰C; T_B = 19⁰C; T_C = 28⁰C; m_A = m_B = m_C = m; temperature of mixture A and B is 16⁰C; temperature of mixture B and C is 23⁰C.

For mixture A and B: temperature of mixture, T = 16⁰C.

→Heat gain by A = heat loss by B

Or, m_A C_A (T – T_A) = m_B C_B (T_B – T)

Or, m C_A (16 – 12) = m C_B (19 – 16)

Or, 4 C_A = 3 C_B

For mixture B and C: temperature of mixture, T = 23⁰C.

→Heat gain by B = heat loss by C

Or, m_B C_B (T – T_B) = m_C C_C (T_C – T)

Or, m C_B (23 – 19) = m C_C (28 – 23)

Or, 4 C_B = 5 C_C

For mixture A and C: temperature of mixture, T.

→Heat gain by A = heat loss by C

Or, m_A C_A (T – T_A) = m_C C_C (T_C – T)

Or, m (3/4) C_B (T – 12) = m (4/5) C_B (28 – T)

Or, (3/4) (T – 12) = (4/5) (28 – T)

Or, 15 T – 180 = 448 – 16 T

Or, T = 628/31 = 20.3⁰C.

4. Four 2 cm * 2 cm * 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg/m³, density of the drink = 1000 kg/m³, specific heat capacity of the drink = 4200 J/kg-K, latent heat of fusion of ice = 3.4 * 10⁵ J/kg.

Sol:

Given: Density of ice, ρ_i = 900 kg/m³; specific heat capacity of ice, C_i = 2108 J/kg-K; density of the drink, ρ_d = 1000 kg/m³; specific heat capacity of the drink, C_d = 4200 J/kg-K; latent heat of fusion of ice, L = 3.4 * 10⁵ J/kg; volume of ice, V_i = 4 * 2³ = 32 cm³ = 32 * 10^-6 m³; volume of drink, V_d = 200 ml = 2 * 10^-4 m³.

Mass of the ice, m_i = ρ_i V_i = 900 * 32 * 10^-6 = 288 * 10^-4 kg

Mass of the drink, m_d = ρ_d V_d = 1000 * 2 * 10^-4 = 0.2 kg

Heat required to melt ice completely is

→ Q_m = m_iL = 288 * 10^-4 * 3.4 * 10⁵ = 9792 J.

Heat release when drink comes from 10⁰ C to 0⁰ C is

→ Q_d = m_dC_dΔT = 0.2 * 4200 * (10 – 0) = 8400 J.

Since, Q_d is greater than Q_m. So complete ice will not melt.

(a) The temperature of the drink when thermal equilibrium is attained is 0⁰ C.

(b) Let the amount ice melt be m.

→ Latent heat of melted ice = Heat loss by the drink

Or, mL = 8400

Or, 3.4 * 10⁵ * m = 8400

Or, m = 8400/ (3.4 * 10⁵) = 0.025 kg = 25 g.

5. Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5°C. Specific heat capacity of water = 4200 J/kg-°C and latent heat of vaporization of water = 2.27 * 10⁶ J/kg.

Sol:

Given: mass of water, m_w = 10 kg; C_w = 4200 J/kg-°C; latent heat of vaporization, L = 2.27 * 10⁶ J/kg; rate at which water comes, m’ = 0.2 g/s = 0.2 * 10^-3 kg/s; ΔT = 5⁰ C. let time = t.

According to the question,

→ Heat loss by the water = latent heat of vaporization of water

Or, m_w * C_w * ΔT = m’ * t * L

Or, 10 * 4200 * 5 = 0.2 * 10^-3 * 2.27 * 10⁶ * t

Or, t = 462.55 s = 7.7 min.

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