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Sol. 1. T base 1 = 15° t base 2 = 17°c ∆t = t base 2 – t base 1 = 17 – 15 = 2°C = 2 + 273 = 275 K m base v = 100 g = 0.1 kg m base w = 200 g = 0.2 kg cu base g = 420 J/kg–k W base g = 4200 J/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 = 1764 J. (c)dQ = 0, dU = – dw = 1764. [since dw = –ve work done on the system]
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