Answer to Question #112027 in Differential Equations for Suresh singh

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

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

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

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

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