Question

A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Answer 1

According to Newton’s law of cooling, we have:

`- (dT)/(dt) = K(T - T_0)`

`(dT)/(K(T-T_0)) = -kdt` ...(i)

Where,

Temperature of the body = T

Temperature of the surroundings = T₀ = 20°C

K is a constant

Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

Integrating equation (i), we get:

`int_50^80 (dt)/(K(T - T_0)) = -int_0^300Kdt`

`[log_e(T-T_0)]_50^80 = -K[t]_0^300`

`2.3026/K log_10 (80-20)/(50-20) = -300`

`2.3026/K =log_10 2 = -300`   ....(ii)

The temperature of the body falls from 60°C to 30°C in time = t^’

Hence, we get:

`2.3026/K log_10  (60-20)/(30-20) = -t`

`-2.3026/t log_10 4 = K` ...(iii)

Equating equations (ii) and (iii), we get:

`-2.3026/t log_10 4 = (-2.3026)/300 log_10 2`

:.t  = 300 x 2 = 600 s = 10 min

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

Answer 2

According to Newton's law of cooling, the rate of cooling is proportional to the difference in temperature.`

Here Average of `80 ^@C` and `50 ^@C = 65 ^@C`

Temperature of surroundings = `20^@C`

:. Difference = `65 - 20 = 45 ^@C`

Under these condition. the body cools `30^@C` in time 5 minutes

`:.  "Change in temp"/"Time" = K triangleT` or `30/5 = K xx 45^@`  .. (1)

The average of  `60^@C` and `30^@` is `45^@C` which is `25^@C`(45 - 20) above the room temperature anf the bodycppls by `30^@C`(60 - 30) in time t (say)

`:. 30/t =  K xx 25`   ...(ii)

Where K is same for this situation as for the original.

Dividing equation i by ii we get

`="30/5"/"30/t" = (Kxx45)/(Kxx25)`

or `t/5 = 9/5`

`=> t = 9 min`