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

B, C

Explanation

Given `M = 2kg`
`Delta theta = 4^@C s_u = 4200J/kgK.`
`p_0 = 999.9kg/m^3`
`p_1 = 1000kg/m^3,P = 10^6kpa.`
`Let internal energy = Delta V`
`DeltaQ = DeltaU+DeltaW`
`rArr ms Delta theta = DeltaU+P(V_0-V_4)`
`rArr 2xx4200xx4 = delta U+10^2(V_0)-(V_4)`
`rArr 33600 = DeltaU+10^5((m)/(p_0)-(m)/(p_4))`
`rArr = Delta U +10^5xx0.0000002`ltbr.`rArr 33600 = Delta U +0.02`
`Delta U = (33600-0.02)J`