The specific heat of alcohol is about half th | vedclass

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(B) The final equilibrium temperature $T$ of a mixture of two substances with masses $m_1, m_2$,specific heats $S_1, S_2$,and initial temperatures $T_

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closer to $T_W$ than $T_A$

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(B) The final equilibrium temperature $T$ of a mixture of two substances with masses $m_1, m_2$,specific heats $S_1, S_2$,and initial temperatures $T_1, T_2$ is given by:$T = \frac{m_1 S_1 T_1 + m_2 S_2 T_2}{m_1 S_1 + m_2 S_2}$Given that the masses are identical $(m_A = m_W = m)$:$T = \frac{m S_A T_A + m S_W T_W}{m S_A + m S_W} = \frac{S_A T_A + S_W T_W}{S_A + S_W}$Given that the specific heat of alcohol is half that of water $(S_A = \frac{1}{2} S_W)$:$T = \frac{(\frac{1}{2} S_W) T_A + S_W T_W}{\frac{1}{2} S_W + S_W} = \frac{\frac{1}{2} T_A + T_W}{\frac{3}{2}} = \frac{T_A + 2 T_W}{3}$$T = \frac{1}{3} T_A + \frac{2}{3} T_W$Since the weight of $T_W$ is twice that of $T_A$ in the final expression,the final temperature $T$ is closer to $T_W$ than to $T_A$.

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