Question

A cup of water is cooling. Its initial temperature is 100𝑜𝐶. After 3 minutes its temperature is 80𝑜𝐶. The temperature 𝑇 of the water, measured in ℃, is modelled by 𝑑𝑇 𝑑𝑡 = −𝑘(𝑇 − 25) where 𝑡 is the time elapsed in minutes. (i) Show that 𝑇 − 𝐴𝑒 −𝑘𝑡 − 25 = 0, where 𝐴 and 𝑘 are appropriate constants, (ii) Find the temperature of the water after 5 minutes.

Answer 1

(i) Let $T(t)=$ the temperature of a cup of water  at time $t.$

The differential equation involving $T$

$$\dfrac{dT}{dt}=-k(T-1000), k>0$$ $$\dfrac{dT}{T-25}=-kdt$$

Integrate

$$\int \dfrac{dT}{T-250}=-\int kdt$$

$$\ln(|T-25|)=-kt+\ln A$$

$$T-25=Ae^{-kt}$$

$$T-Ae^{-kt}-25=0$$

(ii)

Given $T(0)=100\degree C, T(3)=80\degree C$

$$100=25+Ae^{-k(0)}=>A=75$$

$$T(t)=25+75e^{-kt}$$

$$80=25+75e^{-k(3)}$$

$$e^{3k}=\dfrac{75}{55}$$

$$3k=\ln (\dfrac{15}{11})$$

$$k=\dfrac{1}{3}\ln (\dfrac{15}{11})$$

$$T(t)=25+75(\dfrac{11}{15})^{t/3}$$

$$T(5)=25+75(\dfrac{11}{15})^{5/3}$$

$$T(5)\approx69.7\degree C$$