Newton's Law of Cooling Problem: Calculate Coffee Cooling Time to 30°C

Question

A cup of coffee has cooled from 94°C to 50°C after 12 minutes in a room at 24°C. How long will it take to cool to 30°C? (Round to three decimal places as needed.)

Accepted Answer

26.650 minutes

Answer

Step 1: Recall Newton's Law of Cooling formula: $T(t) = T_s + (T_0 - T_s)e^{-kt}$, where $T(t)$ is the temperature at time $t$, $T_s$ is the surrounding temperature, $T_0$ is the initial temperature, $k$ is the cooling constant, and $t$ is time in minutes. Step 2: Identify known values: $T_0 = 94°C$, $T_s = 24°C$, $T(12) = 50°C$, $t=12$ minutes. First solve for $k$. Step 3: Substitute values into the formula to find $k$: $50 = 24 + (94 - 24)e^{-12k}$. Simplify: $50-24=70e^{-12k} \rightarrow 26=70e^{-12k}$. Divide both sides by 70: $\frac{26}{70}=e^{-12k} \rightarrow 0.3714=e^{-12k}$. Step 4: Take the natural logarithm of both sides: $\ln(0.3714) = -12k \rightarrow -0.992 = -12k$. Solve for $k$: $k = \frac{0.992}{12} \approx 0.0827$. Step 5: Now find the time $t$ when $T(t)=30°C$. Substitute into the formula: $30 = 24 + (94-24)e^{-0.0827t}$. Simplify: $6=70e^{-0.0827t} \rightarrow \frac{6}{70}=e^{-0.0827t} \rightarrow 0.0857=e^{-0.0827t}$. Step 6: Take the natural logarithm of both sides: $\ln(0.0857) = -0.0827t \rightarrow -2.457 = -0.0827t$. Solve for $t$: $t=\frac{2.457}{0.0827} \approx 26.650$ minutes.