Question

Only 10 kJ heat energy is available to heat two 1 kg metal balls A and B of specific heats 2 Jkg-1K-1 and 4 Jkg-1K-1. If the temperature change is same (25 ˚C to 75 ˚C), then heat absorbed by A and B respectively are:

• A. 3.33 kJ, 6.67 kJ​
• B. 6.67 kJ, 3.33 kJ​
• C. 5 kJ, 5 kJ​
• D. 4 kJ, 6 kJ​

Answer 1

The correct option is A 3.33 kJ, 6.67 kJ​

Total energy available = 10 kJ
Specific heat capacity of ball A ($c_A$) = 2 Jkg-1K-1
Specific heat capacity of ball B ($c_B$) = 4 Jkg-1K-1
Mass of A = Mass of B = 1 kg
Assume out of 10 kJ available energy, X kJ is absorbed by A.
So, the remaining energy (10-X) kJ is absorbed by B.
Change in temperature for both the balls is same i.e. from 25 ˚C to 75 ˚C
So, ∆T = 50 ˚C
Heat absorbed or released by a body due to a temperature difference is given by
Q = m c ∆T
Heat absorbed by A = X kJ
X (kJ) = (1 kg) (2 Jkg-1K-1) (50 ˚C) ………..(1)
Heat absorbed by B = (10-X) kJ
(10-X) (kJ) = (1 kg) (4 Jkg-1K-1) (50 ˚C) ……..(2)
Dividing (1) by (2) we get,

$\frac{X}{10-X}=\frac{1\times 2\times 50}{1\times 4\times 50}$

200 X = 100 (10 – X)
300 X = 1000
X= 3.33 kJ
10-X = 6.67 kJ
So, metal ball A absorbs 3.33 kJ of energy and metal ball B absorbs 6.67 kJ of energy.