The specific heat capacity of a metal at low temperature (T) is given as: Cp (kjK-1 kg-1) = 32 ((T/400))3 A 100 gram vessel of this metal is to be cooled from 20°K to 4°K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is :

Question

The specific heat capacity of a metal at low temperature (T) is given as: Cp (kjK-1 kg-1) = 32 ((T/400))3 A 100 gram vessel of this metal is to be cooled from 20°K to 4°K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is : (A) greater than 0.148 kJ (B) between 0.148 kJ and 0.028 kJ (C) less than 0.028 kJ (D) equal to 0.002 kJ

Tardigrade Admin · Accepted Answer

Q = ∫ mcdT =∫ limits420 0.1× 32 ×((T3/4003))dt ≈ 0.002 k J.

Q. The specific heat capacity of a metal at low temperature (T) is given as : Cp​(kjK−1kg−1)=32(400T​)3 A 100 gram vessel of this metal is to be cooled from 20∘K to 4∘K by a special refrigerator operating at room temperature (27∘C). The amount of work required to cool the vessel is :

greater than 0.148 kJ

between 0.148 kJ and 0.028 kJ

less than 0.028 kJ

equal to 0.002 kJ

Q=∫mcdT =20∫4​0.1×32×(4003T3​)dt ≈0.002kJ.

$Q = \int m c d T$
$= 20 \int 4 0.1 \times 32 \times (\frac{T ^{3}}{40 0 ^{3}}) d t$
$\approx 0.002 k J .$