Question 4.4.8.4: Using newton’s law of Cooling An object is heated to 100°C (......
Using newton’s law of Cooling
An object is heated to 100°C (degrees Celsius) and is then allowed to cool in a room whose air temperature is 30°C.
(a) If the temperature of the object is 80°C after 5 minutes, when will its temperature be 50°C?
(b) Determine the elapsed time before the temperature of the object is 35°C.
(c) What do you notice about the temperature as time passes?
(a) Using formula (4) with T = 30 and u_{0} = 100, the temperature u(t) (in degrees Celsius) of the object at time t (in minutes) is
u(t) = T + (u_0 – T)e^{kt} ~~~~ k \lt 0 (4)
u(t)\,=\,30\,+\,(100\,-\,30)\,e^{k t}=30\,+\,70e^{k t}where k is a negative constant. To find k, use the fact that u = 80 when t = 5. Then
u(t)\,=\,30\,+\,70e^{k t}\textstyle{80}=30+70e^{k(5)} u(5) = 80.
50=70 e^{5k} Simplify.
e^{5k}={\frac{50}{70}} Solve for e^{5k}
5k=\ln{\frac{5}{7}} Take ln of both sides.
k=\frac{1}{5}\ln\frac{5}{7}\approx-0.0673 Solve for k
Formula (4), therefore, becomes
u(t)=30+70e^{-0.0673t} (5)
Find t when u = 50°C.
50\,=\,30\,+\,70e^{-0.0673t}20\,=\,70e^{-0.0673t} Simplify
e^{-0.0673t}={\frac{20}{70}}-0.0673t=\ln\frac{2}{7} Take ln of both sides.
t={\frac{\ln{\frac{2}{7}}}{-0.0673}}\approx18.6\operatorname*{minutes} Solve for t.
The temperature of the object will be 50°C after about 18.6 minutes, or 18 minutes, 36 seconds.
(b) Use equation (5) to find t when u = 35°C.
35=30+70e^{-0.0673t}5=70e^{-0.0673t} Simplify
e^{-0.0673t}=\frac{5}{70}-0.0673t=\ln{\frac{5}{70}} Take ln of both sides.
t={\frac{\ln{\frac{5}{70}}}{-0.0673}}\approx39.2~{\mathrm{minutes}} Solve for t.
The object will reach a temperature of 35°C after about 39.2 minutes.
(c) Look at equation (5). As t increases, the exponent – 0.0673t becomes unbounded in the negative direction. As a result, the value of e^{-0.0673t} approaches zero, so the value of u, the temperature of the object, approaches 30°C, the air temperature of the room.