The temperature of 0.05 kg of air is raised
by 1°C at constant volume. Calculate the
increase in intermal energy (Cv = 718 J kg C)
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Explanation:
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Answered by
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
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