Answer to Question #122476 in Differential Equations for Joseph Se

Question

1. A small metal bar, whose initial temperature was 30° C, is dropped into a large container of boiling water. How long will it take the bar to reach 80° C if it is known that its temperature increases 2° during the first second? (The boiling temperature for water is 100° C. Round your answer to one decimal place.)

____ sec

How long will it take the bar to reach 96° C? (Round your answer to one decimal place.)

_____ sec

Answer 1

$T^{\prime}=k\left(T-T_{m}\right) ~~~~\text{(Newton’s Law of Cooling)} \\[1 em] \therefore \int \frac{d T}{T-T_{m}}=\int k d t\\[1 em] \therefore \ln \left|T-T_{m}\right|=k t+C_{1} \rightarrow T=T_{m}+c e^{k t} \\[1 em] \because T(0)=30 \quad, \quad T_{m}=100\\[1 em] \therefore 30=100+c e^{0} \rightarrow c=-70 \\[1 em] \therefore T=100-70 e^{k t}\\[1 em] \text{The temperature increased} ~ 2^{\circ} ~ \text{at} ~~ t=1 ~ sec\\[1 em] \therefore 30+2=100-70 e^{k} \\[1 em] \therefore e^{k}=\frac{100-32}{70}=\frac{34}{35} \rightarrow k=\ln \left(\frac{34}{35}\right)=-0.029 \\[1 em] \therefore T=100-70 e^{-0.029 t}\\[1 em] \text{At T}=80 \rightarrow 80=100-70 e^{-0.029 t}\\[1 em] \therefore t=\frac{\ln ((100-80) / 70)}{-0.029}=43.2 \text { sec } \\[1 em] \text{At T}=96 \rightarrow 96=100-70 e^{-0.029 t}\\[1 em] \therefore t=\frac{\ln ((100-96) / 70)}{-0.029}=98.7 \text { sec } \\[1 em]$