Multiple choice

A thermally insulated vessel contains two liquid at temperature $T_1$ and $T_2$ respectively and specific heats $C_1$ and $C_2$ separated by a non-conducting partition. The partition is removed and the difference between the initial temperature of one of the liquids and the temperature T established in the vessel turns out to be equal to half the difference between the initial temperature of the liquids. Determine the ratio $\displaystyle \dfrac{m_1}{m_2}$ (masses of the liquids).

  1. $\displaystyle \dfrac{C_1}{C_2}$
  2. $\displaystyle \dfrac{C_2}{C_1}$
  3. $\displaystyle \dfrac{C_2 + C_1}{C_2 - C_1}$
  4. $\displaystyle \dfrac{C_2 - C_1}{C_2 + C_1}$
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B Correct answer
Explanation

Heat lost = Heat gained: m1*C1*(T1-T) = m2*C2*(T-T2). Given T-T2 = (T1-T2)/2, then T = (T1+T2)/2. Substituting T into the heat equation: m1*C1*(T1 - (T1+T2)/2) = m2*C2*((T1+T2)/2 - T2). This simplifies to m1*C1*(T1-T2)/2 = m2*C2*(T1-T2)/2. Thus, m1/m2 = C2/C1.

AI explanation

By the principle of calorimetry, the heat lost by one liquid equals the heat gained by the other, so m1C1(T1 - T) = m2C2(T - T2). Rearranging the ratio of masses gives m1 divided by m2 equals C2(T - T2) divided by C1(T1 - T). Since the temperature difference between the initial temperature of one liquid and the final temperature is half the total initial difference, we have T - T2 = (T1 - T2) divided by 2 and T1 - T = (T1 - T2) divided by 2, making their ratio exactly 1. Therefore, the ratio m1 to m2 equals C2 divided by C1.

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