A) \[2.4\times {{10}^{4}}\,cal\]
B) \[6\times {{10}^{4}}\,cal\]
C) \[1.2\times {{10}^{4}}\,cal\]
D) \[4.8\times {{10}^{4}}\,cal\]
Correct Answer: C
Solution :
| Key Idea: The heat converted to work is the amount of heat that remains after going through sink. |
| From the relation |
| \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] |
| Given, \[{{Q}_{1}}=6\times {{10}^{4}}\,cal,\] |
| \[{{T}_{1}}=227+273=500\,K\] |
| \[{{T}_{2}}=127+273=400\,K\] |
| \[\therefore \] \[\frac{{{Q}_{2}}}{6\times {{10}^{4}}}=\frac{400}{500}\] |
| \[\Rightarrow \] \[{{Q}_{2}}=\frac{4}{5}\times 6\times {{10}^{4}}\] |
| \[=4.8\times {{10}^{4}}\,cal\] |
| Now, heat converted to work |
| \[={{Q}_{1}}-{{Q}_{2}}\] |
| \[=6.0\times {{10}^{4}}-4.8\times {{10}^{4}}\] |
| \[=1.2\times {{10}^{4}}\,cal\] |
| Note: Carnot cycle consists of following four stages: |
| (i) Isothermal expansion |
| (ii) Adiabatic expansion |
| (iii) Isothermal compression |
| (iv) Adiabatic compression |
| After doing the calculations for different processes, we achieve the reaction |
| \[\frac{{{Q}_{2}}}{Q{{ & }_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] |
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