question_answer

A carnot freezer takes heat from water at \[^{o}C\] inside it and rejects it to the room at a temperature of\[{{27}^{o}}C\]. The latent heat of ice is\[336\times {{10}^{3}}J\,k{{g}^{-1}}\]. If 5 kg of water at \[{{0}^{o}}C\]is converted into ice at \[{{0}^{o}}C\]by the freezer, then the energy consumed by the freezer is close to:   JEE Main Online Paper (Held On 10 April 2016)

A) \[1.68\times {{10}^{6}}J\]                                               

B) \[1.71\times {{10}^{7}}J\]                

C) \[1.51\times {{10}^{5}}J\]                                               

D) 1\[1.67\times {{10}^{5}}J\]

Correct Answer: D

Solution :

                                                    heat required to freeze 5 kg water \[=5\times 336\times {{10}^{3}}\] \[=168\times {{10}^{3}}\]Joule \[\Rightarrow {{Q}_{1}}=1680\,\,KJ\] for carnot's cycle \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\frac{{{Q}_{2}}}{1680}=\frac{300}{273}\] \[{{Q}_{2}}=1680\times \frac{300}{273}KJ\] \[W-{{Q}_{2}}-Q{{ & }_{1}}\] \[=1680\left( \frac{300}{273}-1 \right)\] \[=\frac{1680\times 27}{273}\times {{10}^{3}}J\] \[=166.15\times {{10}^{3}}J\] \[=166.15\times {{10}^{5}}KJ\]