Exercise · ZW-NEWTONS-LAW-OF-COOLING-005 · ★★☆ Intermediate
A body at \(100\,^\circ\)C in surroundings at \(20\,^\circ\)C drops to \(60\,^\circ\)C in \(5\) min. What is \(k\)?
Choose one option.
Show answer
- A. \(k=\dfrac{\ln 2}{5}\)
Explanation
From \(60=20+80e^{-5k}\) we get \(e^{-5k}=\tfrac12\), so \(k=\ln2/5\).
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
- ZW-NEWTONS-LAW-OF-COOLING-001 Solve T'=-0.1 (T-20) with T(0)=80.
- ZW-NEWTONS-LAW-OF-COOLING-002 A coffee at 90 ^∘C is left in a room at 22 ^∘C. What temperature does it approach over time?
- ZW-NEWTONS-LAW-OF-COOLING-003 A body at temperature T sits in surroundings at 15 ^∘C. Which equation expresses Newton's law with…
- ZW-NEWTONS-LAW-OF-COOLING-004 With k=0.2 and T_a=20, what is T' when T=50?
- ZW-NEWTONS-LAW-OF-COOLING-006 Using the data from the previous exercise (from 100 ^∘C to 60 ^∘C in 5 min, surroundings 20 ^∘C), w…
- ZW-NEWTONS-LAW-OF-COOLING-007 A drink at 5 ^∘C is taken out into surroundings at 25 ^∘C. What is the sign of T' at the start, and…
- ZW-NEWTONS-LAW-OF-COOLING-008 With k=0.25 per minute, in how many minutes does the difference T-T_a fall to 1/e of its initial va…
- ZW-NEWTONS-LAW-OF-COOLING-009 Surroundings at 20 ^∘C. A body goes from 90 ^∘C to 55 ^∘C in 10 min. When does it reach 30 ^∘C?
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