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Newton's law of cooling

Problem 6.394 · hard

An object at 100° is placed in a room at 25°. After 15 minutes it has cooled to 85°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 30°.

T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
\[ \frac{d}{d t} \left(25 + 75 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{15}}\right) = - 5 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{15}} \ln{\left(\frac{5}{4} \right)} \]
The model satisfies the cooling law.✓ Proved
\[ 85 \]
k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/15 matches the reading at t = 15.✓ Proved
\[ \frac{317}{5} \]
T(45).✓ Proved
\[ \frac{15 \ln{\left(15 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(15^{\frac{15}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]
Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved

Answer \( T(45) = \frac{317}{5} \approx 63.40^\circ,\quad t = \ln{\left(15^{\frac{15}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 182.04\text{ min} \)

Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	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: fail (error) — The solution fails to explicitly state the value of the target temperature (30°) used to derive the time t in step 5. While the algebraic form is correct, the sentence describing the step is incomplete without specifying that 'target' refers to 30°, making the logical connection to the final numerical answer opaque.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to explicitly state the value of the target temperature (30°) used to derive the time t in step 5. While the algebraic form is correct, the sentence describing the step is incomplete without specifying that 'target' refers to 30°, making the logical connection to the final numerical answer opaque.
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to explicitly state the value of the constant k or the specific function T(t) used to derive the final answers, making the transition from the general law to the specific numerical results opaque. Additionally, the final expression for t is unnecessarily complex and obscures the standard form t = ln(15)/k, though algebraically correct.
gpt-oss:20b: pass 2026-10-12

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 2026-10-12 with SymPy 1.14.0.