0.5 kg of lemon squash at 30 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 C?

Calorimetry Numerical Question: 8 [reference: Selina ICSE class 10 Physics]

0.5 kg 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 kg-1 K-1.

Solution:

Change in temperature = 30 – 5= 25^o C = 25 K

m = 0.5 kg

c = 4200 J kg-1 K-1

△Q = mc△T

△Q = 0.5 × 4200 × 25

△Q = 52500 J …………….. [1]

P = heat flow per second = 30 J/S

Say, time required = t sec.

t = △Q / P

t = 52500 / 30 s

t = 1750 s

t = 29 min 10 sec