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}.


Learn more about this topic:

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What are Heating and Cooling Curves?

from College Chemistry: Help and Review

Chapter 6 / Lesson 5
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