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QPage 47: Q3: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q3 Solution: The temperatures of equal masses of three different liquids A,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the liquids as if they have the same specific heat and simply average the initial temperatures; instead, set up the heat‑balance $mC_i(T_{\,final}-T_i)=mC_j(T_j-T_{,final})$ and use the given mixed‑temperature data to relate the $C$ values before solving for the unknown temperature.

The temperatures of equal masses of three different liquids A, B and C are 12�C, 19�C and 28�C respectively. The temperature when A and B are mixed is 16�C, and when B and C are mixed, it is 23�C. What will be the temperature when A and C are mixed?

Step-by-Step Verified Solution

**Answer:** The temperature when A and C are mixed is 20.3�C. **Step-by-Step Explanation:** TA=12�CT_A = 12�\text{C}; TB=19�CT_B = 19�\text{C}; TC=28�CT_C = 28�\text{C}; mA=mB=mC=mm_A = m_B = m_C = m; temperature of mixture A and B is 16�C; B and C is 23�C. For mixture A and B:
mCA(16−12)=mCB(19−16)m C_A (16 - 12) = m C_B (19 - 16)
4CA=3CB... (1)4 C_A = 3 C_B \quad \text{... (1)}
For mixture B and C:
mCB(23−19)=mCC(28−23)m C_B (23 - 19) = m C_C (28 - 23)
4CB=5CC... (2)4 C_B = 5 C_C \quad \text{... (2)}
When A and C are mixed, if TT is the common temperature:
mCA(T−12)=mCC(28−T)m C_A (T - 12) = m C_C (28 - T)
Substituting CA=34CBC_A = \frac{3}{4}C_B and CC=45CBC_C = \frac{4}{5}C_B:
34CB(T−12)=45CB(28−T)\frac{3}{4}C_B(T - 12) = \frac{4}{5}C_B(28 - T)
15T−180=448−16T15T - 180 = 448 - 16T
T=20.258�C≈20.3�CT = 20.258�\text{C} \approx 20.3�\text{C}
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