Please send me the answer of above q attached with explain it by each step

Dear student

Given M = 2 kg 2t = 4°c Sw = 4200 J/Kg–k f0 = 999.9 kg/m^3 f4 = 1000 kg/m^3 P = 10^5 Pa. Net internal energy = dv dQ = DU + dw ⇒ ms?Q∅ = dU + P(v^0 – v^4) ⇒ 2 × 4200 × 4 = dU + 10^5(m – m) ⇒ 33600 = dU + 10^5 (m/V base 0 – m/v base 4) = dU + 10^5(0.0020002 – 0.002) = dU + 10^5 0.0000002 ⇒ 33600 = du + 0.02 ⇒ du = (33600 – 0.02) J