Question Statement

Ayesha was preparing a pizza in a baking oven. She observed that the temperature of the cooked pizza was $150^{\circ }\mathrm{C}$. Four minutes after removing it from the oven, the temperature of the pizza was $90^{\circ }\mathrm{C}$. How long will it take to cool off to a temperature of $40^{\circ }\mathrm{C}$ if the room temperature is $20^{\circ }\mathrm{C}$?

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Background and Explanation

This problem is solved using Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient (surrounding) temperature. This relationship is expressed as a first-order separable differential equation.

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Solution

According to Newton's Law of Cooling:

$\frac{dT}{dt}=-k\left(T-T_r\right)$

Where:

• $T$ is the temperature of the pizza at any time $t$.
• $T_r$ is the room temperature.
• $t$ is the time elapsed.
• $k$ is the cooling constant.

Step 1: Set up and solve the differential equation

Given that the room temperature $T_r=20^{\circ }\mathrm{C}$, we substitute this into equation (1):

$\frac{dT}{dt}=-k(T-20)$

We separate the variables to prepare for integration:

$\frac{dT}{T-20}=-kdt$

Now, integrate both sides:

$\int \frac{1}{T-20}dT=-k\int dt$ $\ln (T-20)=-kt+A$

Step 2: Find the integration constant $A$

We use the initial condition. At $t=0$ (the moment the pizza is removed from the oven), the temperature $T=150^{\circ }\mathrm{C}$:

$\ln (150-20)=-k(0)+A$ $\ln 130=0+A$ $A=\ln (130)$

Substitute $A$ back into equation (2):

$\ln (T-20)=-kt+\ln (130)$ $\ln (T-20)-\ln 130=-kt$

Using the property of logarithms $\ln (a)-\ln (b)=\ln (\frac{a}{b})$:

$\ln \left(\frac{T-20}{130}\right)=-kt$

Step 3: Find the cooling constant $k$

We are given that at $t=4$ minutes, the temperature $T=90^{\circ }\mathrm{C}$. Substitute these values into equation (3):

$\ln \left(\frac{90-20}{130}\right)=-k(4)$ $\ln \left(\frac{70}{130}\right)=-4k$ $\ln \left(\frac{7}{13}\right)=-4k$

Based on the provided data calculations: $-0.619\approx -4k$ $k=\frac{0.619}{4}$ $k=0.155$

Now our specific equation for this pizza is: $\ln \left(\frac{T-20}{130}\right)=-(0.155)t$

Step 4: Calculate the time for $T=40^{\circ }\mathrm{C}$

To find how long it takes to reach $40^{\circ }\mathrm{C}$, we set $T=40$ and solve for $t$:

$\ln \left(\frac{40-20}{130}\right)=-(0.155)t$ $\ln \left(\frac{20}{130}\right)=-(0.155)t$ $\ln \left(\frac{2}{13}\right)=-0.155t$

Using the calculated value for the natural log: $-1.872=-0.155t$ $t=\frac{1.872}{0.155}$ $t=12.07$ $t\approx 12\text{ minutes}$

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Key Formulas or Methods Used

• Newton's Law of Cooling: $\frac{dT}{dt}=-k(T-T_r)$
• Separation of Variables: A method for solving differential equations by grouping all $T$ terms on one side and $t$ terms on the other.
• Logarithmic Integration: $\int \frac{1}{u}du=\ln |u|+C$
• Logarithm Properties: $\ln (a)-\ln (b)=\ln (\frac{a}{b})$

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Summary of Steps

1. Substitute the room temperature ($20^{\circ }\mathrm{C}$) into the Newton's Law of Cooling formula.
2. Solve the differential equation using integration to get the general logarithmic form.
3. Use the initial temperature ($150^{\circ }\mathrm{C}$ at $t=0$) to find the constant $A$.
4. Use the second temperature reading ($90^{\circ }\mathrm{C}$ at $t=4$) to calculate the cooling constant $k$.
5. Plug the target temperature ($40^{\circ }\mathrm{C}$) into the resulting equation and solve for the time $t$.