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:
\(\int_{80}^{50}\frac{dT}{K}(T-T_0)\) = - \(\int_{300}^{0}K\,dt\) 
log_e (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.