Question

A cup of coffee has cooled from 95°C to 50°C after 15 minutes in a room at
20°C. How long will it take to cool to 40°C?

Answer 1

Answer:

21.6 min

Note: This answer was rounded to three significant figures.

Explanation:

To solve this problem, we can use 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 temperature. This can be modeled by the equation:

$\boxed{ \begin{array}{ccc} \text{\underline{Newton's Law of Cooling:}} \\\\ \dfrac{dT}{dt} = -k(T - T_a) \\\\ \text{Where:} \\ \bullet \ \frac{dT}{dt} \ \text{is the rate of change of temperature} \\ \bullet \ k \ \text{is the cooling constant} \\ \bullet \ T \ \text{is the temperature of the object} \\ \bullet \ T_{a} \ \text{is the ambient temperature} \end{array} }$

This differential equation can be solved to give an exponential decay model for the temperature of the coffee over time:

$\boxed{ \begin{array}{ccc} \text{\underline{Temperature Over Time (Newton's Law of Cooling):}} \\\\ T(t) = T_a + (T_0 - T_a)e^{-kt} \\\\ \text{Where:} \\ \bullet \ T(t) \ \text{is the temperature of the object at time } t \\ \bullet \ T_a \ \text{is the ambient temperature} \\ \bullet \ T_0 \ \text{is the initial temperature of the object} \\ \bullet \ k \ \text{is the cooling constant} \\ \bullet \ t \ \text{is the time} \end{array} }$

We are given:

• T_a = 20°C
• T₀ = 95°C
• T(15 min) = 50°C

We can use the above information to determine the cooling constant 'k':

$\Longrightarrow 50 = 20 + (95 - 20)e^{-k(15)}\\\\\\\\\Longrightarrow 50 = 20 + 75e^{-15k}\\\\\\\\\Longrightarrow 30 = 75e^{-15k}\\\\\\\\\Longrightarrow \dfrac{2}{5} = e^{-15k}$

$\Longrightarrow \ln\left(\dfrac{2}{5}\right) = -15k\\\\\\\\\Longrightarrow k=\dfrac{\ln\left(\dfrac{2}{5}\right)}{-15}\\\\\\\\\therefore k \approx 0.06109$

We want to determine 't' when T(t) = 40°C:

$T(t) = T_a + (T_0 - T_a)e^{-0.06109k}\\\\\\\\\Longrightarrow 40 = 20 + (95 - 20)e^{-0.06109t}$

$\Longrightarrow 40 = 20 + 75e^{-0.06109t}\\\\\\\\\Longrightarrow 20 = 75e^{-0.06109t}\\\\\\\\\Longrightarrow \dfrac{4}{15} = e^{-0.06109t}\\\\\\\\\Longrightarrow \ln\left(\dfrac{4}{15}\right) = -0.06109t$

$\Longrightarrow t = \dfrac{\ln\left(\dfrac{4}{15}\right)}{-0.06109}$

$\therefore t \approx \boxed{21.6 \text{ min}}$

Thus, it will take approximately 21.6 min to cool the cup of coffee to 40°C.