0.5 kg of lemon squash at 30⁰C is placed in a refrigerator which can remove heat at an average rate of 30 Js^-1. How long will it take to cool the lemon squash to 5⁰C?(Sp. heat capacity of lemon squash = 4200 J kg^{-1 o}C^-1.)

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Solution

$Change in temperature of lemon squash = 30 - 5 = 25 ° C Heat lost by lemon squash, Q = m \times C \times Δ T Q = 0.5 \times 4200 \times 25 = 52500 J Rate at which heat is removed is 30 {Js}^{- 1} \frac{Q}{t} = 30 {Js}^{- 1} \frac{52500 J}{t} = 30 {Js}^{- 1} t = \frac{52500 J}{30 {Js}^{- 1}} = 1750 sec = 29.2 min$

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0.5 k g

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

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

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0.5 kg of lemon squash at 300C is placed in a refrigerator which can remove heat at an average rate of 30 Js-1. How long will it take to cool the lemon squash to 50C?(Sp. heat capacity of lemon squash = 4200 J kg-1 oC-1.)

Change in temperature of lemon squash = 30 − 5 = 25°CHeat lost by lemon squash, Q = m × C × ΔTQ = 0.5 × 4200 × 25 = 52500 JRate at which heat is removed is 30 Js−1Qt = 30Js−152500 Jt = 30Js−1t = 52500 J30Js−1 = 1750 sec = 29.2 min

0.5 kg of lemon squash at 30⁰C is placed in a refrigerator which can remove heat at an average rate of 30 Js^-1. How long will it take to cool the lemon squash to 5⁰C?(Sp. heat capacity of lemon squash = 4200 J kg^{-1 o}C^-1.)