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In two glasses there is liquid material of the same type, the volume of one is $$V_1$$, its temperature is $$T_1$$, the volume of the other is $$V_2$$ and its temperature is $$T_2$$. We mix the liquids and wait while the temperature settles. What is the common temperature if we neglect any heat losses? What would be the common temperature if the volume of first liquid is halved and the second liquid doubled.
Correct Answer: 0
The liquids are identical, so mass is proportional to volume and the specific heat cancels. Heat lost by the hotter liquid equals heat gained by the colder one, $$V_1(T_1 - T) = V_2(T - T_2)$$, giving $$T = \frac{V_1T_1 + V_2T_2}{V_1 + V_2}$$. With volumes $$\frac{V_1}{2}$$ and $$2V_2$$ the same rule gives $$T' = \frac{\frac{V_1}{2}T_1 + 2V_2T_2}{\frac{V_1}{2} + 2V_2} = \frac{V_1T_1 + 4V_2T_2}{V_1 + 4V_2}$$.
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