Using newton’s law of Cooling An object is heated to 100°C (degrees Celsius) and is then

Question

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?

Accepted Answer

(a) Using formula (4) with T = 30 and [latex]u_{0}[/latex] = 100, the temperature u(t) (in degrees Celsius) of the object at time t (in minutes) is

[latex] u(t) = T + (u_0 - T)e^{kt} ~~~~ k \lt 0 [/latex] (4)

[latex]u(t)\,=\,30\,+\,(100\,-\,30)\,e^{k t}=30\,+\,70e^{k t}[/latex]

where k is a negative constant. To find k, use the fact that u = 80 when t = 5. Then

[latex]u(t)\,=\,30\,+\,70e^{k t}[/latex]

[latex] extstyle{80}=30+70e^{k(5)}[/latex] u(5) = 80.

[latex]50=70 e^{5k}[/latex] Simplify.

[latex]e^{5k}={\frac{50}{70}}[/latex] Solve for [latex]e^{5k}[/latex]

[latex]5k=\ln{\frac{5}{7}}[/latex] Take ln of both sides.

[latex]k=\frac{1}{5}\ln\frac{5}{7}\approx-0.0673[/latex] Solve for k

Formula (4), therefore, becomes

[latex]u(t)=30+70e^{-0.0673t}[/latex] (5)

Find t when u = 50°C.

[latex]50\,=\,30\,+\,70e^{-0.0673t}[/latex]

[latex]20\,=\,70e^{-0.0673t}[/latex] Simplify

[latex]e^{-0.0673t}={\frac{20}{70}}[/latex]

[latex]-0.0673t=\ln\frac{2}{7}[/latex] Take ln of both sides.

[latex]t={\frac{\ln{\frac{2}{7}}}{-0.0673}}\approx18.6\operatorname*{minutes}[/latex] 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.

[latex]35=30+70e^{-0.0673t}[/latex]

[latex]5=70e^{-0.0673t}[/latex] Simplify

[latex]e^{-0.0673t}=\frac{5}{70}[/latex]

[latex]-0.0673t=\ln{\frac{5}{70}}[/latex] Take ln of both sides.

[latex]t={\frac{\ln{\frac{5}{70}}}{-0.0673}}\approx39.2~{\mathrm{minutes}}[/latex] 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 [latex]e^{-0.0673t}[/latex] approaches zero, so the value of u, the temperature of the object, approaches 30°C, the air temperature of the room.

Question 4.4.8.4: Using newton’s law of Cooling An object is heated to 100°C (......

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