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Newton's law of cooling

Problem 6.234 · hard

An object at 50° is placed in a room at 0°. After 10 minutes it has cooled to 40°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 10°.

T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
Reviewed
\[ \frac{d}{d t} 50 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{10}} = - 5 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{10}} \ln{\left(\frac{5}{4} \right)} \]
The model satisfies the cooling law.✓ Proved
\[ 40 \]
k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/10 matches the reading at t = 10.✓ Proved
\[ \frac{128}{5} \]
T(30).✓ Proved
\[ \frac{10 \ln{\left(5 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(5^{\frac{10}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]
Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved

Answer \( T(30) = \frac{128}{5} \approx 25.60^\circ,\quad t = \ln{\left(5^{\frac{10}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 72.13\text{ min} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	k fitted by a root-finder, then the cooling law integrated numerically

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly sets up Newton's Law of Cooling, determines the constant k, and calculates the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically redundant.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly sets up Newton's Law of Cooling, determines the constant k, and calculates the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically redundant.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly sets up Newton's Law of Cooling, determines the constant k, and computes the required values. The final expression for time is algebraically equivalent to the standard form, though written in a redundant way.
gpt-oss:20b: pass 2026-10-04

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/newtons_cooling, checked 2026-10-04 with SymPy 1.14.0.