### Math Notes

Subjects

#### Differential Equation Solutions

##### Topics || Problems

A thermometer reading is $$18^oF$$ is brought into a room where the temperature is $$70^oF$$; 1 min later the thermometer reading is $$31^oF$$. Determine the temperature reading as a function of time and, in particular, find the temperature reading 5 min after the thermometer is first brought into the room.
 Room Temperature $$70^oF$$ Time Thermometer Reading $$0 ~min$$ $$18^oF$$ $$1~min$$ $$31^oF$$

Temperature reading as a function of time.

Fom Newtons Law of Cooling $$\frac{dT_b}{dt} = k (T_b - T_s)$$, where $$T_b$$ is the temperature of the body, $$T_s$$ is the temperature of the sorroundings and k is a constant.

$$\frac{dT_b}{T_b-T_s} = kdt$$, $$T_s$$ is the room temperature and assumed to be constant.

$$\ln (T_b - T_s) + ln C= kt$$

At $$t=0 ~T_s = 70^oF, ~ T_b = 18^oF$$

$$\ln (18-70) + ln C= 0$$

$$\ln C = -ln (-52)$$

$$\ln(T_b - T_s) - \ln(-52) = kt$$

At $$t = 1~ T_s = 31^oF$$

$$\ln(31-70)-\ln(-52) = k$$

$$k = \ln \frac{-39}{-52}$$

$$k = \ln \frac{3}{4} = -0.29$$

$$\ln(T_b - 70) - \ln(-52) = \ln (\frac{3}{4})t$$

$$\frac{T_b-70}{-52} = e^{-0.29t}$$

$$T_b = 70 -52 e^{-0.29t}$$ Answer

At $$t = 5$$

$$T_b = 70-52e^(-.29x5)$$

$$T_b \approx 57.66^oF \approx 58^oF$$ Answer