250,g of water and an equal volume of alcohol | vedclass

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(C) According to Newton's law of cooling,the rate of loss of heat is proportional to the temperature difference between the body and the surroundings.

Vedclass · Accepted Answer

$0.62$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of loss of heat is proportional to the temperature difference between the body and the surroundings. Since the temperature range ($60^{\circ}C$ to $55^{\circ}C$) is the same for both,the average rate of heat loss is approximately constant.Let $m_w = 250\,g$ (mass of water),$m_a = 200\,g$ (mass of alcohol),$W = 10\,g$ (water equivalent of calorimeter),and $s$ be the specific heat of alcohol. The specific heat of water $s_w = 1\,	ext{cal/g}^{\circ}C$.Heat lost by water and calorimeter: $Q_w = (m_w + W) s_w \Delta T = (250 + 10) 	imes 1 	imes 5 = 1300\,	ext{cal}$.Rate of heat loss for water: $R_w = \frac{Q_w}{t_w} = \frac{1300}{130} = 10\,	ext{cal/s}$.Heat lost by alcohol and calorimeter: $Q_a = (m_a s + W) \Delta T = (200s + 10) 	imes 5 = 1000s + 50\,	ext{cal}$.Rate of heat loss for alcohol: $R_a = \frac{Q_a}{t_a} = \frac{1000s + 50}{67}$.Since the rate of heat loss is the same in both cases $(R_w = R_a)$:$10 = \frac{1000s + 50}{67}$$670 = 1000s + 50$$1000s = 620$$s = 0.62\,	ext{cal/g}^{\circ}C$.

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