Exercise · ZW-NEWTONS-LAW-OF-COOLING-004 · ★☆☆ Basic
With \(k=0.2\) and \(T_a=20\), what is \(T'\) when \(T=50\)?
Choose one option.
Show answer
- C. \(-6\)
Explanation
\(T'=-0.2\,(50-20)=-0.2\cdot30=-6\).
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.
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- 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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