Exercise · ZW-NEWTONS-LAW-OF-COOLING-009 · ★★★ Advanced

Surroundings at \(20\,^\circ\)C. A body goes from \(90\,^\circ\)C to \(55\,^\circ\)C in \(10\) min. When does it reach \(30\,^\circ\)C?

Choose one option.

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  • B. \(t=10\log_2 7\approx28.1\) min

Explanation

The difference goes from \(70\) to \(35\) in \(10\) min, so \(e^{-10k}=\tfrac12\). For \(30\,^\circ\)C the difference is \(10=70e^{-kt}\): \(e^{kt}=7\), \(t=\ln7/k=10\ln7/\ln2\).

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Review the concept: Newton's law of cooling

States that the temperature \(T\) of a body, with initial value \(T_0\), changes at a rate proportional to its difference from the ambient temperature \(T_a\), with constant \(k>0\).

\[T'=-k\,(T-T_a)\;\Longrightarrow\; T(t)=T_a+(T_0-T_a)\,e^{-kt}\]

More practice problems: newton's law of cooling

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