Question

0.5 kg of lemon squash at 30°C 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 5°C?(Sp. heat capacity of lemon squash = 4200 J kg^{−1 o}C⁻¹.)

Answer 1

Change in temperature of lemon squash,

ΔT = 30 − 5

= 25°C

Heat lost by lemon squash,

Q = mcΔT

= 0.5 × 4200 × 25

= 52500
Rate at which heat is removed is 30 Js⁻¹,

`Q/t = 30  Js^-1`

`(52500  J)/t = 30  Js^-1`

t = `(52500  J)/ (30  Js^-1)`

= 1750 sec

= 29.2 min