Question

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.)

Answer 1

Change in temperature of lemon squash = 30 - 5 = 25°C

Heat lost by lemon squash, Q = m x C x ΔT

Q = 0.5 x 4200 x 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