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-T0)
$\frac{dT}{K(T-T_0)}$ = -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:
$\int_{80}^{50}\frac{dT}{K}(T-T_0)$ = - $\int_{300}^{0}K\,dt$
loge (T-T0)]5080 = - K[t]0 300
$\frac{2.3026}{K}$ log10 $\frac{80-20}{50-20}$ = - 300
$\frac{2.3026}{K}$ log102 = - 300
$\frac{2.3026}{300}$ log10 2 = K ......... (ii)
The temperature of the body falls from 60°C to 30°C in time = t’
Hence, we get:
$\frac{2.3026}{K}$ log10 $\frac{60-20}{30-20}$= -t
$-\frac{2.3026}{K}$= log10 4 = K .......... (iii)
Equating equations (ii) and (iii), we get:
$-\frac{2.3026}{t}$ log10 4 = $\frac{-2.3026}{300}$ log10 2
∴ t = 300x2 = 600 s = 10 min
Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.