Multiple choice

Calculate the initial temperature of liquid $A$ of mass $400$g having a specific heat capacity of $\displaystyle 2.4J{ kg }^{ -1 }{ K }^{ -1 }$ when it is mixed with liquid $B$ of mass $750$g having a specific heat capacity of $\displaystyle 1.6J{ kg }^{ -1 }{ K }^{ -1 }$ at $\displaystyle { 40 }^{ \circ }C$. The final temperature of the mixture becomes $\displaystyle { 25 }^{ \circ }C$.

  1. $\displaystyle { 15 }^{ \circ }C$
  2. $\displaystyle { 7.5 }^{ \circ }C$
  3. $\displaystyle { 4.2 }^{ \circ }C$
  4. $\displaystyle { 6.25 }^{ \circ }C$
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D Correct answer
Explanation

Using the principle of calorimetry, the heat gained by liquid A equals the heat lost by liquid B. This gives the equation m_A * c_A * (T_f - T_A) = m_B * c_B * (T_B - T_f). Substituting the given values, we get 400 * 2.4 * (25 - T_A) = 750 * 1.6 * (40 - 25), which simplifies to T_A = 6.25 degrees Celsius.

AI explanation

Using the method of mixtures formula, m_A * c_A * (T_A - T_f) = m_B * c_B * (T_f - T_B), we set the mass in grams and specific heats to find the lost and gained heat. This gives 400 * 2.4 * (T_A - 25) = 750 * 1.6 * (25 - 40), which simplifies to 960T_A - 24000 = -18000. Solving for the initial temperature of liquid A yields 960T_A = 6000, giving T_A = 6.25 degrees Celsius.

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