The specific heat capacity of a metal at low temperature (T) is given as C_p=32left(frac{ T }{400}right)^{3};kJ,k ^{-1}, kg ^{-1}. A 100; g vessel of

Q: 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

c Specific heat at low is $C_{P}=32\left(\frac{T}{400}\right)^{3}$ $Q=\int m \cdot c . d T=\int_{20}^{4} \frac{100}{1000} \times 32\left(\frac{T}{400}\right)^{3} d T$ $=\frac{32}{10} \times \frac{1}{(400)^{3}}\left(\frac{T^{4}}{4}\right)^{3}$ $=\frac{32}{10 \times(400)^{3}} \times \frac{1}{4}\left(20^{4}-4^{4}\right)$ $=\frac{32}{10 \times(400)^{3}} \times \frac{1}{4} \times(160000-256)$ $=0.002 W$ $\beta=\frac{T_{2}}{T_{1}-T_{2}}=\frac{Q_{3}}{W}$ $\Rightarrow \frac{20}{300-20}=\frac{0.2}{W} \Rightarrow W=0.028 kJ$ $\Rightarrow \frac{4}{300-4}=\frac{0.002}{W} \Rightarrow W=0.0148 kJ$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

Aequal to $0.002\; KJ$
B greater than $0.148\; KJ$
Cbetween $0.148\; KJ$ and $0.028 KJ$
Dless than $0.028\; KJ$

AIEEE 2011, Diffcult

STD 11 Science
NEET
STD 11 - 11. thermodynamics
Physics

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