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A system is provided `50xx10^(3)J` energy and work done on the system is `100J.` The change in internal energy is `:`A. `50kJ`B. `50.1kJ`C. `150J`D. `50J` |
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Answer» Correct Answer - B `q=DeltaU+(-W)` or `DeltaU=q+W` Here `W` is work done on the system `=50xx10^(3)+100=50100J` `=50.1kJ` |
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