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 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:
Change in temperature = 30 – 5= 25o 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