The specific heat capacity of a metal at low temperature (T) is g… - Vidyadip

MCQ

The specific heat capacity of a metal at low temperature $(T)$ is given as $C_p=32\left(\frac{ T }{400}\right)^{3}\;kJ\,k ^{-1}\, kg ^{-1}$. A $100\; g$ vessel of this metal is to be cooled from $20 \;K$ to $4\; K$ by a special refrigerator operating at room temperature $27^\circ c$). The amount of work required to cool the vessel is

A
equal to $0.002\; KJ$
B
greater than $0.148\; KJ$
✓
between $0.148\; KJ$ and $0.028 KJ$
D
less than $0.028\; KJ$

✓

Answer

Correct option: C.

between $0.148\; KJ$ and $0.028 KJ$

c
Specific heat at low is

$C_{P}=32\left(\frac{T}{400}\right)^{3}$

$Q=\int m \cdot c . d T=\int_{20}^{4} \frac{100}{1000} \times 32\left(\frac{T}{400}\right)^{3} d T$

$=\frac{32}{10} \times \frac{1}{(400)^{3}}\left(\frac{T^{4}}{4}\right)^{3}$

$=\frac{32}{10 \times(400)^{3}} \times \frac{1}{4}\left(20^{4}-4^{4}\right)$

$=\frac{32}{10 \times(400)^{3}} \times \frac{1}{4} \times(160000-256)$

$=0.002 W$

$\beta=\frac{T_{2}}{T_{1}-T_{2}}=\frac{Q_{3}}{W}$

$\Rightarrow \frac{20}{300-20}=\frac{0.2}{W} \Rightarrow W=0.028 kJ$

$\Rightarrow \frac{4}{300-4}=\frac{0.002}{W} \Rightarrow W=0.0148 kJ$

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