# If c = 100, k = 0.1, and t is time in minutes, how long will it take a hot cup of coffee to cool...

## Question:

If c = 100, k = 0.1, and t is time in minutes, how long will it take a hot cup of coffee to cool to a temperature of {eq}25^{\circ} {/eq}C in a room at {eq}20^{\circ}{/eq} C?

## Newton?s Law of Cooling:

Newton?s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the ambient.

## Answer and Explanation:

Thanks to the Newton's Law of Cooling and the condition of the problem, we can write the equation:

{eq}\frac{{dT}}{{dt}} = - k\left( {T - {T_a}} \right)\\ T\left( 0 \right) = {T_0}\\ \\ k = 0.1,\;{T_a} = 20\\ \\ \frac{{dT}}{{dt}} = - 0.1\left( {T - 20} \right) {/eq}

Solving the differential equation:

{eq}\frac{{dT}}{{dt}} = - 0.1\left( {T - 20} \right)\\ \frac{{dT}}{{T - 20}} = - 0.1dt\\ \int {\frac{{dT}}{{T - 20}}} = \int { - 0.1dt} \\ \ln \left( {T - 20} \right) = - 0.1t + m\\ T - 20 = {e^{ - 0.1t + m}}\\ T - 20 = {e^{ - 0.1t}}{e^m}\\ T = 20 + {e^{ - 0.1t}}c\\ \\ c = 100\\ T\left( t \right) = 20 + 100{e^{ - 0.1t}} {/eq}

Calculating the time we are looking for, we have:

{eq}T\left( t \right) = 25\\ 25 = 20 + 100{e^{ - 0.1t}}\\ 5 = 100{e^{ - 0.1t}}\\ {e^{ - 0.1t}} = \frac{5}{{100}} = 0.05\\ - 0.1t = \ln \left( {0.05} \right)\\ t = \frac{{\ln \left( {0.05} \right)}}{{ - 0.1}} \approx 30\min {/eq}

So the hot cup of coffee needs 30 min to cool to a temperature of {eq}25^{\circ} {/eq}.