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A cooking vessel on a slow burner contains 5 kg of water and an unknown mass of ice in equilibrium at $0^{0} C .$ at time t = 0. The temperature of the mixture is measured at various times and the result is plotted as shown in figure. During the first 50 min the mixture remains at $0^{0} C .$ From 50 min to 60 min, the temperature increases to $2^{0} C .$ Neglecting the heat capacity of the vessel, the initial mass of the ice is (Assume rate of heat given by the burner is constant)[take 1cal. Is 4.2J]

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$\frac{10}{7} k g$

$\frac{5}{7} k g$

$\frac{5}{4} k g$

$\frac{5}{8} k g$

answer is B.

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Detailed Solution

Let ‘m’ be the mass of ice.

Rate of heat given by the burner is constant. In the first 50 min

$\frac{d Q}{d t} = \frac{m L}{t_{3}} = \frac{m k g \times (80 \times 4.2 \times 10^{3}) J / k g}{(50 \min)} ... (i)$

From 50 min to 60 min

$\frac{d Q}{d t} = \frac{(m + 5) S_{H_{2} O} Δ θ}{t_{2}} = \frac{(m + 5) k g (4.2 \times 10^{3}) J / k g \times 2^{0} C}{10 \min} ..... (i i)$

From eq. (i) and (ii), $\frac{80 m}{50} = \frac{2 (m + 5)}{10}, 7 m = 5 \Rightarrow m = \frac{5}{7} k g = 0.7 k g$

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