Water temperature cooling

The water in the container has a temperature of t1 = 80◦C, the temperature around the container is t2 = 15◦C. The dependence of temperature t on time τ (in minutes) can be expressed approximately by the formula:

t = t2 +(t1 −t2)·e^(−0.05·τ)

Calculate the temperature of the water
a) after 5 minutes;
b) after 1 hour.

Final Answer:

t₃ =  65.6221 °C
t₄ =  18.2362 °C

Step-by-step explanation:

$$\begin{gathered} t_1=80\text{°C} \\ {t}_2=15\text{°C} \\ {\tau }_1=5\text{min} \\ {k}_1=-0.05\cdot {\tau }_1=-0.05\cdot 5=-\frac{1}{4}=-0.25 \\ {t}_3=t_2+(t_1-t_2)\cdot e^{k_1}=15+(80-15)\cdot e^{(-0.25)}=65.6221\text{°C} \end{gathered}$$

$$\begin{gathered} {\tau }_2=1\text{h}\to \text{min}=1\cdot 60\text{min}=60\text{min} \\ {k}_2=-0.05\cdot {\tau }_2=-0.05\cdot 60=-3 \\ {t}_4=t_2+(t_1-t_2)\cdot e^{k_2}=15+(80-15)\cdot e^{(-3)}=18.2362\text{°C} \end{gathered}$$

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