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

AnSwer :-

Heat required to change the temperature of the vessel by a small amount dT; dQ = mc$\sf _p$dT

Total heat required

$\sf Q = m \int \limits_{20} ^{4} 32 \bigg( \dfrac{</p><p>T}{400} \bigg)^{3} d</p><p>T = 0.001996kJ$

work done required to maintain the temperature of sink T₂

W = Q₁ - Q₂

$\sf W = \bigg(\dfrac{Q_{1} - Q₂}{Q₂}\bigg)Q₂ = \bigg( \dfrac{T_1 - T_2}{T_2} \bigg)Q₂$

For, T₂ = 20K

$\sf W_1 = \bigg( \dfrac{300 - 20}{20} \bigg)0.001996 = 0.028kJ$

For, T₂ = 4K

$\sf W_2 = \bigg( \dfrac{300 - 4}{4} \bigg) 0.001996 = 0.148kJ$

$\therefore$ the work required to cool the vessel from 20K to 4K is

W₂ - W₁ = 0.148 - 0.028 = 0.12KJ.

As the temperature is changed from 20K to 4K work done required will be more than W₁ but less than W₂

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

#sanvi....