t1 = 15°c t2 = 17°c
Δt = t2 – t1 = 17 – 15 = 2°C = 2 + 273 = 275 K
mv = 100g = 0.1kg
mw = 200g = 0.2kg
cug = 420J/kg–k
Wg = 4200J/kg–k
(a) The heat transferred to the liquid vessel system is 0. The internal heat is shared in between the vessel and water.
(b) Work done on the system = Heat produced unit => dw = 100 × 10–3 × 420 × 2 + 200 × 10–3 × 4200 × 2 = 84 + 84 × 20 = 84 × 21 = 1764J.
(c) dQ = 0, dU = – dw = 1764. [since dw = –ve work done on the system]