Question

The rate of change of temperature of an object is proportional to the difference between

the temperature of the object and its environment. A glass of hot water at a temperature

of 70ºC is kept in a room which is at a temperature of 30 ºC. If after 3 minutes the

temperature of the water is 50 ºC, what will be its temperature after 5 minutes?

Answer 1

The rate $\frac{dt}{d\tau}$ of the change of temperature of the water is given by

$$\frac{dt}{d\tau}=k(t-t_{env}),$$

where $k$ is the proportionality coefficient,

$t$ - current temperature of the water,

$t_{env}$ - environmental temperature.

Let us find the coefficient by solving the equation with initial and final conditions defined by the first case.

$$\frac{dt}{t-t_{env}}=kd\tau,$$

$$\int_{70}^{50}\frac{dt}{t-t_{env}}=k\int_0^3d\tau,$$

$$ln(t-t_{env})\Big\rvert_{70}^{50}=k\tau\Big\rvert_0^3,$$

$$k=\frac{1}{3}ln\Bigg(\frac{70-30}{50-30}\Bigg)=0.23105.$$

Substitution the conditions defined by the second case lets us find the final temperature $t_f$ of the water after 5 minutes of the cooldown

$$ln(t-t_{env})\Big\rvert_{70}^{t_f}=k\tau\Big\rvert_0^5,$$

$$ln\Bigg(\frac{70-30}{t_f-30}\Bigg)=5k,$$

$$\frac{40}{t_f-30}=e^{5k},$$

$$t_f=\frac{40}{e^{5\cdot0.23105}}+30=42.60^oC.$$