Question

The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A , where T is the water temperature, A is the room temperature, and k is a positive constant. If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, how long (to the nearest minute) will it take the water to cool to 60°C?

Answer 1

Answer:7.96 min

Explanation:

Given

time taken by water to cool from $90^{\circ}C$ to $85^{\circ}C$ is 1 min

Ambient Temperature $T_{\infty }=30^{\circ}C$

According to Newtons law of cooling

$\frac{T-T_{\infty }}{T_i-T_{\infty }}=e^{-kt}$

where T=Final Temperature

$T_i$=Initial Temperature

k=time constant

t=time

$\frac{85-30}{90-30}=e^2{-k\cdot 1}$

$\frac{11}{12}=e^{-k}$

$k=0.08701$

Time taken to cool to $60^{\circ}C$

$\frac{60-30}{90-30}=e^{-kt}$

$\frac{1}{2}=e^{-kt}$

taking Log

$\ln 0.5=-kt$

$t=\frac{0.6931}{0.08701}$

$t=7.96 min$