# Q. 11.20 A body cools from  $\inline 80^{\circ}C$ to $\inline 50^{\circ}C$  in 5 minutes. Calculate the time it takes to cool from $\inline 60^{\circ}C$ to  $\inline 30^{\circ}C.$ The temperature of the surroundings is  $\inline 20^{\circ}C.$

Let a body initially be at temperature T1

Let its final Temperature be T2

Let the surrounding temperature be T0

Let the temperature change in time t.

According to Newton's Law of cooling

$\\-\frac{dT}{dt}=K(T-T_{0})\\ \frac{dT}{T-T_{0}}=-Kdt\\ \int_{T_{1}}^{T_{2}}\frac{dT}{T-T_{0}}=-K\int_{0}^{t}dt\\ \left [ ln(T-T_{0}) \right ]_{T_{1}}^{T_{2}}=-K[t]_{0}^{t}\\ ln\left ( \frac{T_{2}-T_{0}}{T_{1}-T_{0}} \right )=-Kt$

where K is a constant.

We have been given that the body cools from 80 oC to 50 oC in 5 minutes when the surrounding temperature is 20 oC.

T2 = 50 oC

T1 = 80 oC

T0 = 20 oC

t = 5 min = 300 s.

$\\ln\left ( \frac{50-20}{80-20} \right )=-K\times 300\\ K=\frac{ln(2)}{300}\\$

For T1 = 60 oC and T2 = 30 oC we have

$\\ln\left ( \frac{30-20}{60-20} \right )=-\frac{ln2}{300}t\\ t=\frac{ln(4)\times 300}{ln(2)} \\t=\frac{2ln(2)\times 300}{ln(2)}\\ t=600\ s\\ t= 10\ min$

The body will take 10 minutes to cool from 60 oC to 30 oC at the surrounding temperature of 20 oC.

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