Questions: Newton's Law of Cooling says that the rate at which a body cools is proportional to the difference C in temperature between the body and the environment around it. The temperature f(t) of the body at time t in hours after being introduced into an environment having constant temperature T0 is f(t)=T0+Ce^(-kt), where C and k are constants. A cup of coffee with temperature 135°F is placed in a freezer with temperature 0°F. After 10 minutes, the temperature of the coffee is 60°F. Use Newton's Law of Cooling to find the coffee's temperature after 20 minutes. After 20 minutes the coffee will have a temperature of °F. (Round to the nearest integer as needed.)

Solution Steps

We are given the initial temperature of the coffee, the temperature of the environment, and the temperature of the coffee after 10 minutes. We need to find the temperature of the coffee after 20 minutes using Newton's Law of Cooling.

The temperature of the coffee at time \( t \) is given by: \[ f(t) = T_0 + C e^{-kt} \] where:

• \( T_0 = 0^\circ \mathrm{F} \) (temperature of the freezer),
• \( f(0) = 135^\circ \mathrm{F} \) (initial temperature of the coffee),
• \( f(10) = 60^\circ \mathrm{F} \) (temperature after 10 minutes).

First, find \( C \) using the initial condition: \[ f(0) = T_0 + C e^{-k \cdot 0} = 135 \] \[ 0 + C \cdot 1 = 135 \] \[ C = 135 \]

Next, use the condition at 10 minutes to find \( k \): \[ f(10) = T_0 + C e^{-10k} = 60 \] \[ 0 + 135 e^{-10k} = 60 \] \[ e^{-10k} = \frac{60}{135} = \frac{4}{9} \]

Take the natural logarithm of both sides to solve for \( k \): \[ -10k = \ln\left(\frac{4}{9}\right) \] \[ k = -\frac{1}{10} \ln\left(\frac{4}{9}\right) \]

Substitute \( k \) back into the equation to find \( f(20) \): \[ f(20) = 0 + 135 e^{-20k} \] \[ f(20) = 135 \left(\frac{4}{9}\right)^{2} \]

Calculate: \[ \left(\frac{4}{9}\right)^{2} = \frac{16}{81} \] \[ f(20) = 135 \times \frac{16}{81} \approx 26.6667 \]

Final Answer