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Find the change in the internal energy of `2 kg` of water as it heated from `0^(0)C to 4^(0)C`. The specific heat capacity of water is `4200 J kg^(-1) K^(-1)` and its densities at `0^(0)C` and `4^(0)C` are `999.9 kg m^(-3)` and `1000 kg m^(-3)` respectively. atmospheric pressure `=10^(5)` Pa. |
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Answer» Correct Answer - `(33600+ 0.02)J` `m = 2kg` `DeltaT = 4k` `C = 4200 J//kg-K` `dU = ? ` `dQ = dU +dW` `(mCdT) = dU +PdV` `2 xx 4200 xx4 =dU +10^(5) (-dV)` `(33600+0.02) = dU` |
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