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A 1.0 kg bar of copper is heated at atmospheric pressure `(1.01xx10^5N//m^2)`. If its temperature increases from `20^@C` to `20^@C`, calculate the change in its internal energy. `alpha=7.0xx10^-6//^@C`, `rho=8.92xx10^3kg//m^3` and `c=387J//kg-^@C`. |
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Answer» Correct Answer - A::B `V=(m)/(rho)=(1.0)/(8.92xx10^3)` `=1.12xx10^-4m^3` `gamma=3alpha=2.1xx10^-5per^@C` `DeltaV=VgammaDeltatheta` `=(1.12xx10^-4)(2.1xx10^-5)(30)` `=7.056xx10^-8m^3` `W=pDeltaV` `=(1.01xx10^5)(7.056xx10^-8)` `=7.13xx10^-3J` `Q=mcDeltaQ` =(1)(387)(30)` `Delta U=Q-W` `=11609.99287J` |
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