Using Newton’s Law of Cooling A cake removed from the oven has a temperature of 210°F. It is left to cool in a room that has a temperature of 70°F. After 30 minutes, the temperature of the cake is 140°F. a. Use Newton’s Law of Cooling to find a model for the temperature of the cake, T, after t

Question

Using Newton’s Law of Cooling

A cake removed from the oven has a temperature of 210°F. It is left to cool in a room that has a temperature of 70°F. After 30 minutes, the temperature of the cake is 140°F.

a. Use Newton’s Law of Cooling to find a model for the temperature of the cake, T, after t minutes.

b. What is the temperature of the cake after 40 minutes?

c. When will the temperature of the cake be 90°F?

Accepted Answer

a. We use Newton’s Law of Cooling

[latex]T = C + (T_0 - C)e^{kt}[/latex].

When the cake is removed from the oven, its temperature is 210°F. This is its initial temperature: [latex]T_0[/latex] = 210. The constant temperature of the room is 70°F: C = 70. Substitute these values into Newton’s Law of Cooling. Thus, the temperature of the cake, T, in degrees Fahrenheit, at time t, in minutes, is

[latex]T = 70 + (210 - 70)e^{kt} = 70 + 140e^{kt}[/latex].

After 30 minutes, the temperature of the cake is 140°F. This means that when t = 30, T = 140. Substituting these numbers into Newton’s Law of Cooling will enable us to find k, a negative constant.

[latex]T = 70 + 140e^{kt}[/latex] Use Newton’s Law of Cooling from above.

[latex]140 = 70 + 140e^{k·30}[/latex] When t = 30, T = 140. Substitute these numbers into the cooling model.

[latex]70 = 140e^{30k}[/latex] Subtract 70 from both sides.

[latex]e^{30k} =\frac{1}{2}[/latex] Isolate the exponential factor by dividing both sides by 140. We also reversed the sides.

[latex]\ln e^{30k} = \ln(\frac{1}{2})[/latex] Take the natural logarithm on both sides.

[latex]30k = \ln(\frac{1}{2})[/latex] Simplify the left side using [latex]\ln e^x = x[/latex].

[latex]k=\frac{ln(\frac{1}{2})}{30}≈-0.0231[/latex] Divide both sides by 30 and solve for k.

We substitute -0.0231 for k into Newton’s Law of Cooling, [latex]T = 70 + 140e^{kt}[/latex]. The temperature of the cake, T, in degrees Fahrenheit, after t minutes is modeled by

[latex]T = 70 + 140e^{-0.0231t}[/latex].

b. To find the temperature of the cake after 40 minutes, we substitute 40 for t into the cooling model from part (a) and evaluate to find T.

[latex]T = 70 + 140e^{-0.0231(40)} ≈ 126[/latex]

After 40 minutes, the temperature of the cake will be approximately 126°F.

c. To find when the temperature of the cake will be 90°F, we substitute 90 for T into the cooling model from part (a) and solve for t.

[latex]T = 70 + 140e^{-0.0231t}[/latex] This is the cooling model from part (a).

[latex]90 = 70 + 140e^{-0.0231t}[/latex] Substitute 90 for T.

[latex]20 = 140e^{-0.0231t}[/latex] Subtract 70 from both sides.

[latex]e^{-0.0231t}=\frac{1}{7}[/latex] Divide both sides by 140. We also reversed the sides.

[latex]\ln e^{-0.0231t} =\ln (\frac{1}{7})[/latex] Take the natural logarithm on both sides.

[latex]-0.0231t =\ln (\frac{1}{7})[/latex] Simplify the left side using [latex]\ln e^x = x[/latex].

[latex]t=\frac{\ln (\frac{1}{7})}{-0.0231}≈84[/latex] Solve for t by dividing both sides by -0.0231.

The temperature of the cake will be 90°F after approximately 84 minutes.

Question 3.5.4: Using Newton’s Law of Cooling A cake removed from the oven h...

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