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 $g^{- 1} K^{- 1}$ .

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Solution

Given,
Mass of lemon squash, m=0.5 kg
Specific heat capacity of lemon squash= 4200 J k $g^{- 1} K^{- 1}$
Fall in temperature, θ= 30−5=25 deg Celsius
Heat extracted by the refrigerator, H=mCθ
=0.5×4200×25=52500 J

Rate of extraction of heat=30J/s
Time in which lemon squash cools=52500/30=1750 sec or 29.17 min

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