Exercise · ZW-NEWTONS-LAW-OF-COOLING-006 · ★★☆ Intermediate

Using the data from the previous exercise (from \(100\,^\circ\)C to \(60\,^\circ\)C in \(5\) min, surroundings \(20\,^\circ\)C), what is the temperature at \(10\) min?

Choose one option.

Show answer
  • B. \(40\,^\circ\)C

Explanation

The difference from the surroundings halves every \(5\) min: \(80\to40\to20\), so \(T(10)=20+20=40\). It is not linear: dropping \(40\) each time would give \(20\).

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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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