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1 kg of water is heated from `40^(@) C "to" 70^(@)C`,If its volume remains constatn, then the change in internal energy is (specific heat of water = 4148 J `kg^(-1 K^(-1))`A. 2.44x`10^(5)J`B. 1.62x`10^(5)J`C. 1.24xx`10^(5)J`D. 2.62x`10^(5)J` |
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Answer» Correct Answer - C Since volume of water remains constant , then work done `DeltaW=PDV=0` According to first pair of thermodynamics `dQ=dU+dW,dU=dQ=msDeltaT` `=1xx4148xx(70-40)=4148xx30` `=124440 J=1.244xx10^(5)J` |
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