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.

Answer

`(33600+ 0.02)J`

Explanation

`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`