At room temperature of 30 degrees C, a glass of water takes one minute to cool from 80 degrees C...

Question:

At room temperature of 30 degrees C, a glass of water takes one minute to cool from 80 degrees C to 60 degrees C. How long will it take to further cool from 60 degrees C to 50 degrees C?

Newton's cooling law

When any two bodies are brought into contact with different temperatures, a heat flux appears between them, so that if they are in contact for a long time the temperature of both bodies will equalize. A classic example of this is the free cooling, or Newton's cooling, of a body in contact with the environment. According to Newton, this process follows the differential equation:

{eq}\displaystyle \frac{\mathrm{d} T}{\mathrm{d} t}=-k(T-T_e) {/eq}

where {eq}T_e {/eq} is the temperature of the environment where the body is located, and {eq}k {/eq} is a specific constant that depends on several factors.

Answer and Explanation:

Solving this equation in this case:

{eq}\displaystyle \frac{\mathrm{d} T}{\mathrm{d} t}=k(T-T_e)\\ \displaystyle \frac{\mathrm{d} T}{T-T_e}=k\mathrm{d} t\\ T(t)=T_e+Ce^{kt}=30^\circ+Ce^{kt}\\ T(0\,\mathrm{s})=80^\circ=30^\circ+C\Rightarrow C=50^\circ\\ T(t)=30^\circ+50^\circ e^{kt}\\ T(60\,\mathrm{s})=60^\circ=30^\circ+50^\circ e^{k(60\,\mathrm{s})}\\ \displaystyle k=\frac{1}{60\,\mathrm{s}}\ln\left (\frac{60^\circ-30^\circ}{50^\circ} \right )=-8.5\cdot10^{-3}\,\mathrm{s^{-1}}\\ \therefore T(t)=30^\circ+50^\circ e^{(-8.5\cdot10^{-3}\,\mathrm{s^{-1}})t} {/eq}

Let us now calculate the time it takes to reach 50 degrees Celsius:

{eq}50^\circ=30^\circ+50^\circ e^{(-8.5\cdot10^{-3}\,\mathrm{s^{-1}})t}\\ \displaystyle t_{50}=\frac{1}{-8.5\cdot10^{-3}\,\mathrm{s^{-1}}}\ln\left (\frac{50^\circ-30^\circ}{50^\circ} \right )=108\,\mathrm{s}\\ {/eq}

As it took one minute to reach 60 degrees Celsius, then it will take 48 s to go from 60 to 50 degrees Celsius (108 s - 60 s = 48 s).


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