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
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