A body at a temperature of 50° F is placed outdoors where the temperature is 100° F. If after 5 minutes the temperature of the body is 60° F, find (a) how long it will take the body to reach a temperature of 75° F and (b) the temperature of the body after 20 minutes.

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Accepted Answer

Using (7.2) with [latex]T_{m}=100[/latex] (the surrounding medium is the outside air), we have

[latex]\frac{d T}{d t}+k T=100 k[/latex]

This differential equation is linear. Its solution is given in Problem 6.15 as

[latex]T=c e^{-k t}+100[/latex] (1)

Since T = 50 when t = 0,it follows from (1) that [latex]50=c e^{-k(0)}+100, ext { or } c=-50[/latex] . Substituting this value into (1), we obtain

[latex]T=-50 e^{-k t}+100[/latex] (2)

At t =5, we are given that T= 60; hence, from (2), [latex]60=-50 e^{-5 k}+100[/latex] . Solving for k, we obtain

[latex]-40=-50 e^{-5 k} \quad ext { or } \quad k=\frac{-1}{5} \ln \frac{40}{50}=\frac{-1}{5}(-0.223)=0.045[/latex]

Substituting this value into (2), we obtain the temperature of the body at any time t as

[latex]T=-50 e^{-0.045 t}+100[/latex] (3)

(a) We require t when T = 75. Substituting T=75 into (3), we have

[latex]75=-50 e^{-0.045 t}+100 \quad ext { or } \quad e^{-0.045 t}=\frac{1}{2}[/latex]

Solving for t, we find

[latex]-0.045 t=\ln \frac{1}{2} \quad ext { or } \quad t=15.4 min[/latex]

(b) We require T when t = 20. Substituting t = 20 into (3) and then solving for T, we find

[latex]T=-50 e^{(-0.045)(20)}+100=-50(0.41)+100=79.5^{\circ} F[/latex]

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Eq. (7.2) : [latex]\frac{d T}{d t}+k T=k T_{m}[/latex]

Question 7.9: A body at a temperature of 50° F is placed outdoors where th...

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