A piece of alloy weighing 4 kg and at a temperature of 800 K is placed in 4 kg of water at 300 K. If — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Using energy balance, $4(4)(800 - T_f) = 4(1)(T_f - 300) \implies 4(800 - T_f) = T_f - 300 \implies T_f = 700 \text{ K}$. $\Delta S_{\text{alloy}} = 16000 \ln\left(\frac{700}{800}\right) = 16000(\ln 7 - 3\ln 2) = 16000(1.95 - 2.1) = -2400 \text{ cal/K}$. $\Delta S_{\text{water}} = 4000 \ln\left(\frac{700}{300}\right) = 4000(1.95 - 1.1) = 3400 \text{ cal/K}$. Total $\Delta S = 3400 - 2400 = 1000 \text{ cal/K} = +1.0 \text{ kcal/K}$.