A) \[150K\]
B) \[100K\]
C) \[26-{{85}^{o}}C\]
D) \[295K\]
Correct Answer: A
Solution :
We know that work done, \[W={{C}_{\upsilon }}({{T}_{1}}-{{T}_{2}})\] \[W=3kJ,\,\,C\upsilon =20j{{K}^{-1}}\] \[{{T}_{1}}=27+273=300K,{{T}_{2}}=?\] \[3\times 1000=20(300-{{T}_{2}})\] \[\therefore \] \[3000=6000-20{{T}_{2}}\] \[\therefore \] \[{{T}_{2}}=\frac{3000}{20}=150K\]You need to login to perform this action.
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