A) 150 K
B) 100 K
C) \[{{26.85}^{o}}C\]
D) 295 K
Correct Answer: C
Solution :
[a] We know that work done, \[W={{C}_{v}}({{T}_{1}}-{{T}_{2}})\] |
\[3\times 1000=20\,(300-{{T}_{2}})\]; \[\therefore 3000=6000-20\,{{T}_{2}}\] |
\[\therefore {{T}_{2}}=\frac{3000}{20}=150\,K\]. |
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