Newton's law of cooling
Problem 6.320 · hard
An object at 85° is placed in a room at 25°. After 20 minutes it has cooled to 73°. Using Newton's law of cooling, find its temperature after 40 minutes and when it reaches 30°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(25 + 60 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}}\right) = - 3 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} \ln{\left(\frac{5}{4} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 73 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{317}{5} \]T(40).✓ Proved
- \[ \frac{20 \ln{\left(12 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(12^{\frac{20}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(40) = \frac{317}{5} \approx 63.40^\circ,\quad t = \ln{\left(12^{\frac{20}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 222.72\text{ min} \)
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: fail (misleading) — The solution’s sentences do not correctly derive the cooling constant or the time to reach 30°. The step claiming “73 = 73” is meaningless, and the final time formula uses an incorrect expression (involving 12) that does not correspond to the target temperature of 30°. This would mislead a student about how to apply Newton’s law of cooling.qwen3.6:27b-mlx: pass — The solution correctly applies Newton's Law of Cooling, derives the constant k accurately from the given data, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though written in a redundant logarithmic form.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies Newton's Law of Cooling, derives the constant k accurately from the given data, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though written in a redundant logarithmic form.gpt-oss:20b: fail (misleading) 2026-10-08 — The solution’s sentences do not correctly derive the cooling constant or the time to reach 30°. The step claiming “73 = 73” is meaningless, and the final time formula uses an incorrect expression (involving 12) that does not correspond to the target temperature of 30°. This would mislead a student about how to apply Newton’s law of cooling.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies Newton's Law of Cooling, derives the constant k, and calculates the requested values. The final expression for time is algebraically equivalent to the standard form, though written in a redundant logarithmic format.gpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.