The specific heat capacity of a metal at low temperature $(T)$ is given as:
$C_{p} (k J K^{- 1} k g^{- 1}) = 32 (\frac{T}{400})^{3}$
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:

$C_{p} (k J K^{- 1} k g^{- 1}) = 32 (\frac{T}{400})^{3}$

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:

TestHub · Accepted Answer

The amount of heat, which needs to be extracted to reduce temperature from 20 K to 4 K is $Q_{ex} = \int dQ = \int mcdT = \int_{4}^{20} 1 0^{- 1} \times 32 \times (\frac{T}{400})^{3} dT = 0.002 kJ$

Performance of refrigerator at 20 K , $β_{1} = \frac{Q _{e x}}{W _{1}} = \frac{T _{2}}{T _{H} - T _{L}} = \frac{20}{300 - 20} = \frac{1}{14}$ or, $W_{1} = 14 Q_{ex} = 0.028 kJ$

Similarly at 4 K $β_{2} = \frac{Q _{ex}}{W _{2}} = \frac{4}{300 - 4} = \frac{1}{74}$ or,

$W_{2} = 74 Q_{ex} = 0.148 kJ$

So, the work done is between 0.028 kJ and 0.148 kJ .

Physics - Elasticity Question with Solution | TestHub

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