A) equal to 0.002 kJ
B) greater than 0.148 kJ
C) between 0.148 KJ and 0.028 kJ
D) less than 0.028 KJ
Correct Answer: C
Solution :
[c] Heat required to change the temperature of vessel by a small amount dT \[-dQ=m{{C}_{p}}dT\] Total heat required \[-Q=m\int_{20}^{4}{32{{\left( \frac{T}{400} \right)}^{3}}dT}\] \[=\frac{100\times {{10}^{-3}}\times 32}{{{(400)}^{3}}}{{\left[ \frac{{{T}^{4}}}{4} \right]}^{4}}_{20}\] \[\Rightarrow \] \[Q=0.001996\,kJ\] Work done required to maintain the temperature of sink to\[{{T}_{2}}\] \[W={{Q}_{1}}-{{Q}_{2}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{2}}}{{Q}_{2}}=\left( \frac{{{T}_{1}}}{{{T}_{2}}}-1 \right){{Q}_{2}}\] \[\Rightarrow \] \[W=\left( \frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{2}}} \right){{Q}_{2}}\] For \[{{T}_{2}}=20\,K\] \[{{W}_{1}}=\frac{300-20}{20}\times 0.001996=0.028kJ\] For \[{{T}_{2}}=4\,K\] \[{{W}_{2}}=\frac{300-4}{4}\times 0.001996=0.148\,kJ\] As temperature is changing from 20k to 4 k, work done required will be more than \[{{W}_{1}}\] but less than \[{{W}_{2}}.\]
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