More practice questions

4. Q4. A calorimeter has a water equivalent of 20 g. It holds 180 g of water at 25 °C. Then 100 g of water at 85 °C is poured in. Ignoring heat loss, find the final temperature.

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   Treat the calorimeter as 20 g of extra water, so the cold side is 200 g at 25 °C. T = (0.20 × 25 + 0.10 × 85)/0.30 = 13.5/0.30 = 45 °C.

5. Q5. A 0.20 kg metal block at 100 °C is dropped into 0.40 kg of water at 20 °C in a calorimeter of water equivalent 0.02 kg. The final temperature is 24 °C. Find the specific heat of the metal, and say which metal in NCERT Table 10.3 it is closest to.

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   Heat gained = (0.40 + 0.02) × 4186 × 4, about 7032 J. Heat lost = 0.20 × s × 76. So s = 7032/15.2, about 463 J kg^⁻¹ K^⁻¹, closest to iron (450 J kg^⁻¹ K^⁻¹).

6. Q6. In an experiment like NCERT Example 10.3, the hot solid cools a little in the air while it is being moved into the calorimeter. Will the specific heat worked out from the readings be too high or too low?

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   Too low. The solid enters the water cooler than the T₁ used in the calculation, so the water and calorimeter gain less heat than m₁s₁(T₁ − T) predicts. Dividing this smaller heat gained by the over-large m₁(T₁ − T) gives too small a value of s₁.