$0.5 k g$ of lemon squash at $30^{o} C$ is placed in a refrigerator which can remove heat at an average rate of $30 J s^{- 1}$ . How long will it take to cool the lemon squash to $5^{o} C$ ? Specific heat capacity of squash $= 4200 J k g^{- 1} K^{- 1}$ .

29 m i n 10 s

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58 m i n 20 s

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39 m i n 10 s

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None of the above

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Solution

The correct option is A $29 m i n 10 s$
Heat removed by refrigerator, $Q = m c Δ T = 0.5 \times 4200 \times (30 - 5) = 52500$ $J$
Now it can remove heat at the rate of $30 J / s$ , and we need to remove $52500 J$ .
So, time is $\frac{52500}{30} = 1750 s = 29 m i n 10 s e c$

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0.5 kg of lemon squash at $30^{\circ} C$ is placed in a refrigerator which can remove heat at an average rate of 30 J $s^{- 1}$ .How long will it take to cool the lemon squash to $5^{\circ} C$ ? Specific heat capacity of squash=4200 J k $g^{- 1} K^{- 1}$ .

0.5 kg of lemon squash at $30^{0} C$ is placed in a refrigerator which can remove heat at an average rate of 30 $J s^{- 1}$ . How long will it take to cool the lemon squash to $5^{0} C$ ? (Specific heat capacity of squash = 4200 J $K g^{- 1} K^{- 1})$ .