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Sol. ℓ = 1 mm = 10^–3 m m = 10 kg A = 200 cm2 = 2 × 10^–2 m^2 L base vap = 2.27 × 10^6 J/kg K = 0.80 J/m-s-°C dQ = 2.27 × 10^6 × 10, dQ/st = 2.27 * 10^7/10^5 = 2.27 * 10^2 J/s again we know dQ/dt = 0.80 * 2 * 10^2 * (42 - T)/1 * 10^-3 So, 8 * 2 * 10^-3(42 -T)/10^-3 = 2.27 * 10^2 ⇒ 16 * 42 – 16T = 227 ⇒ T = 27.8 = 28°C
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