A calorimeter contains 10 g of water at 20. The temperature fall to 15 in 10 min. When calorimeter contains 20 g of water at 20, it takes 15 min. for the temperature to become 15. The water equivalent at the calorimeter is (a) 50 g (b) 25 g (c) 10 g (d) 5 g

Answer: (c). Same calorimeter and same temperature range, so the rate of heat loss is the same in both cases. ((m₁ + W)Δ T)/(t₁) = ((m₂ + W)Δ T)/(t₂) (10 + W)/(10) = (20 + W)/(15) ⇒ 150 + 15W = 200 + 10W W = 10 g So the answer is (c) 10 g.

A calorimeter contains 10 g of water at 20. The temperature fall to… | MHT CET 2025 (26 Apr, Shift 2)

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Newton's Law of Cooling

A calorimeter contains 10 g of water at $20^\circ\text{C}$ . The temperature fall to $15^\circ\text{C}$ in 10 min. When calorimeter contains 20 g of water at $20^\circ\text{C}$ , it takes 15 min. for the temperature to become $15^\circ\text{C}$ . The water equivalent at the calorimeter is

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Answer: (c)

Same calorimeter and same temperature range, so the rate of heat loss is the same in both cases.
$\frac{(m_1 + W)\Delta T}{t_1} = \frac{(m_2 + W)\Delta T}{t_2}$
$\frac{10 + W}{10} = \frac{20 + W}{15} \Rightarrow 150 + 15W = 200 + 10W$
$W = 10$ g
So the answer is (c) 10 g.

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