Question

The specific heat capacity of a metal at low temperature (t) is given as cp=32(t/400)^3 a 100 gram vessel of this metal is to be cooled from 20k to 4k by a special refrigerator operating at room temperature (27°c).The amount of work required to cool the vessel is

Answer 1

Hello dear,

● Answer- 0.002 kJ

● Explaination-
# Given-
m = 100 g = 0.1 kg
cp = 32(T/400)^3 = 5×10^-7 T^3
T1 = 20 K
T2 = 4 K

# Solution-
Amount of work required is given by,
W = ʃdW.dT
W = ʃ(mcp).dT
W = ʃ(0.1×5×10^-7×T^3).dT
W = 5×10^-8 ʃT^3.dT
W = 5×10^-8 [T^4/4]
W = 5×10^-8 / 4 × (20^4-4^4)
W = 5×10^-8 / 4 × 624 × 4^4
W = 0.002 kJ

Amount of work required to cool the vessel is 0.002 kJ.

Hope this helps you...