0 . 08 kg air is heated at constant volume through 5 ° C . The specific heat of air at constant volume is 0 . 17 kcal kg - 1 ° C - 1 and 1 J = 4 . 18 joule cal - 1 . The change in its internal energy is approximately.

Step-by-step solution

The amount of work done for the given process should be zero, as the process is an isochoric process. Thus, the change in internal energy will be equal to the heat exchanged in the process. Mathematically, Q = ∆ U Now, ∆ U = m s ∆ T . . . 1 From equation (1), it follows that ∆ U = 0 . 08 × ( 170 × 4 . 18 ) × 5 J ≈ 284 J