Exercise · ZW-NEWTONS-LAW-OF-COOLING-007 · ★★☆ Intermediate
A drink at \(5\,^\circ\)C is taken out into surroundings at \(25\,^\circ\)C. What is the sign of \(T'\) at the start, and what happens?
Choose one option.
A \(T'<0\): it cools toward \(5\,^\circ\)C B \(T'<0\): it cools below \(5\,^\circ\)C C \(T'=0\): it stays at \(5\,^\circ\)C D \(T'>0\): it warms toward \(25\,^\circ\)C Show answer D. \(T'>0\): it warms toward \(25\,^\circ\)C← Back to the question
Explanation \(T'=-k(5-25)=20k>0\): since \(T<T_a\), the temperature rises and approaches \(T_a\) without passing it.
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