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 J s⁻¹. How long will it take to cool the lemon squash to 5°C? Specific heat capacity of squash = 4200 J g⁻¹K⁻¹.

Answer 1

Given, 

mass (m) = 0.5 kg

Change in temperature = (30 − 5)°C = 25°C = 25 K.

Specific heat capacity of  squash = 4200 J g⁻¹K⁻¹

ΔQ = mcΔT

ΔQ = 0.5 × 4200 × 25 = 52500 J

Let time taken to remove 52500 J of heat be t.

Now it is given that,

30 J of heat is removed in 1 sec

So, 52500 J of heat ⇒ t =?

`"t" = (Δ"Q")/"P"`

= `1/30 xx 525001`

`= 152500 / 30`

= 1750 s

  t = 29 min 10 sec