Question

When 0.012 kg of air is heated from 10 degrees to 40 degrees at a constant volume, what will be the change in the internal energy? The specific heat of air is 0.256 Kcal/kgC.​

Answer 1

Explanation:

heat = ΔQ=Mass × specific heat×temperature change=15×0.2×5 cal=15 cal

Also, isochoric process implies work done =0

Applying first law, Change in internal energy= ΔQ =15 cal