Question

The change in internal energy in joule when 10g of air is heated
from 30°C to 40°C is (c=0.172 kcal/kg/K J=4200J/kcal)​

Answer 1

Explanation:

ANSWER

Given,

T

1

=30

0

C=30+273=303K

T

2

=40

0

C=40+273=313K

m=10g=10

−2

kg

ΔV=0m

3

C

v

=0.172Kcal

Change in Heat energy,

ΔQ=mC

v

(T

2

−T

1

)

ΔQ=10

−2

×0.172×4200×(313−303)

ΔQ=0.172×42×10

ΔQ=72.24J

Work done, W=P.ΔV=0J

From first law of thermodynamics,

ΔQ=ΔU+W

ΔU=ΔQ

ΔU=72.24J