Question

The air temperature is 50 degree celsius .water cools from 100 degree celsius to 75 degree celsius in 5minutes.how long will it take for the water to cool to 10 degree celsius

Answer 1

Newton’s Law of cooling

$\frac{dT}{dt}=K(T-T_0)$

K=liquid constant

T₀= room temperature

T=temperature of the object

$ΔT=100-75 = 25\;°C \\ Δt=5 \;min \\ T_0=50 \;°C \\ T=\frac{100+75}{2}=87.5\;°C \\ \frac{100-75}{5} = K(\frac{100+75}{2}-50) \\ \frac{100-10}{t} = K(\frac{100+10}{2}-50) \\ 5 = 37.5K \\ \frac{90}{t}=5K \\ K=\frac{5}{37.5} \\ t = \frac{90 \times 37.5}{5 \times 5}=135 \;min$

Answer: 135 min