Question

47.

Five kg of air is heated at constant volume. The temperature of air increases from 300k to 340k. if the specific heat at constant volume is 0.169 k cal/kg.. find the amount of heat

absorved. 1) 40k cal

2) 20 kcal

3) 33.8k cal

4) 10k cal​

Answer 1

Answer:

is it physics or what i am not understanding

Answer 2

Answer:

pls mark BRI list

Explanation:

ANSWER

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