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Home » 0.5 kg of lemon squash at 30o 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 5o C? Specific heat capacity of squash = 4200 J kg-1 K-1.
0.5 kg of lemon squash at 30o 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 5o C? Specific heat capacity of squash = 4200 J kg-1 K-1.
Calorimetry | Class 10 | ICSE | Physics | Selina
0.5 kg of lemon squash at 30o 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 5o C? Specific heat capacity of squash = 4200 J kg-1 K-1.

Solution:

According to the question,

Change in temperature = 30 – 5 = 25 K

We know that the expression for heat energy is => △Q = mc△T

Substituting values, we get

△Q = 0.5 × 4200 × 25 = 52500 J

Heat energy can also be expressed in terms of Power

△Q = P × t

Upon re-arranging, we have => t = △Q / P

thus, t = 52500 / 30 = 1750 s

Therefore, t = 29 min 10 sec

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