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

Answers (1)
S Sayak

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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