Numerical problems

Problem 1

The temperature of a metal ball is 30° C. When an energy of 3000 J is supplied, its temperature raises by 40° C. Calculate its heat capacity.

Solution

Heat capacity, C' = Q / ΔT

Here, Q = 3000 J

ΔT = 40°C - 30°C = 10°C or 10 K

C' = 3000 / 10 = 300 JK-1

The heat capacity of the metal ball is 300 JK-1.

Problem 2

The energy required to raise the temperature of an iron ball by 1 K is 500 JK-1. Calculate the amount of energy required to raise its temperature by 20 K.

Solution

Heat capacity, C' = Q / ΔT

 Q=C'×ΔT

Here, C' = 500 JK-1

 ΔT=20K

 Q = 500 × 20 = 10000 J.

The amount of heat energy required is 10000 J.

Problem 3

An energy of 84000 J is required to raise the temperature of 2 kg of water from 60° C to 70° C. Calculate the specific heat capacity of water.

Solution

Specific heat capacity, C = Q / m × ΔT

Here, Q = 84000 J

 m = 2 kg

 ΔT = 70° C – 60° C = 10° C or 10 K

C = 84000 / 2×10 = 4200 J kg-1 K-1

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

Problem 4

The specific heat capacity of a metal is J kg-1K-1. Calculate the amount of heat energy required to raise the temperature of 500 gram of the metal from 125° C to 325° C.

Solution

Specific heat capacity, C = Q / m × ΔT

 Q = C × m × ΔT

Here, C = 160 J kg K-1

m = 500 g = 0.5 kg

ΔT = 325° C – 125° C = 200° C or 200 K

 = 160 × 0.5 × 200 = 16000 J.

The amount of heat energy required is 16000 J.

1. An iron ball requires 1000 J of heat to raise its temperature by 20°C. Calculate the heat capacity of the ball.

Solution :

Heat capacity C’ = Q/∆T

Here, A = 1000 J

T = 20°C − 0°C = 20°C = 20K

C = 1000/20 = 50 JK−1

The heat capacity of the ball = 50 JK−1

2. The heat capacity of the vessel of mass 100 kg is 8000 J/°K. Find its specific heat capacity.

Solution :

Specific heat capacity, C = Q  / [ m × ∆T]

Here, m = 100 kg

Heat capacity = Q / ∆T = 8000 J/°C = 8000 J/K

C = Q  / [ m × ∆T] = 100 × 8000 J = 8,00,000 JKg-1K-1