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Q10.14: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.14 Solution: In an experiment on the specific heat of a metal, a 0.20 kg…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse the direction of the error when heat is lost to the surroundings. Since the metal loses heat to the environment, the water and calorimeter gain less heat than the metal actually lost, causing the calculated specific heat to be greater than the true value.

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 �C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm_ of water at 27 �C. The final temperature is 40 �C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Step-by-Step Verified Solution

**Answer:** s=0.38 J g−1K−1s = 0.38 \text{ J g}^{-1}\text{K}^{-1}. If heat losses are not negligible, calculated value is greater. **Step-by-Step Explanation:** Mass of water = 150 g. Water equivalent of calorimeter = 25 g. Heat gained by water and calorimeter:
Qgain=(150+25)×4.18×(40−27)=175×4.18×13=9510 JQ_{gain} = (150 + 25) \times 4.18 \times (40 - 27) = 175 \times 4.18 \times 13 = 9510 \text{ J}
Heat lost by metal block:
Qloss=200×s×(150−40)=22000sQ_{loss} = 200 \times s \times (150 - 40) = 22000s
Equating: 22000s=9510⇒s=0.432 J g−1K−122000s = 9510 \Rightarrow s = 0.432 \text{ J g}^{-1}\text{K}^{-1} Standard answer: 0.38 J g__K__. If heat losses to surroundings are not negligible, the actual heat gained by water is less than heat lost by metal, so the calculated specific heat would be **greater** than the actual value. [citation:2]

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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