Newton's law of cooling
Problem 6.359 · hard
An object at 82° is placed in a room at 22°. After 5 minutes it has cooled to 52°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 27°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(22 + 60 e^{- \frac{t \ln{\left(2 \right)}}{5}}\right) = - 12 e^{- \frac{t \ln{\left(2 \right)}}{5}} \ln{\left(2 \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 52 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(2)/5 matches the reading at t = 5.✓ Proved
- \[ \frac{367}{16} \]T(30).✓ Proved
- \[ \frac{5 \ln{\left(12 \right)}}{\ln{\left(2 \right)}} = \ln{\left(12^{\frac{5}{\ln{\left(2 \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{367}{16} \approx 22.94^\circ,\quad t = \ln{\left(12^{\frac{5}{\ln{\left(2 \right)}}} \right)} \approx 17.92\text{ min} \)
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: passqwen3.6:27b-mlx: fail (error) — The solution calculates the time to reach 27°C as approximately 17.92 minutes, which is less than the 5 minutes it took to cool to 52°C. Since the object is cooling, it cannot reach a lower temperature (27°C) before it reaches a higher temperature (52°C). The error stems from using the wrong logarithmic identity in line 5: t = ln(12)/k, not ln(12^(1/k)).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution calculates the time to reach 27°C as approximately 17.92 minutes, which is less than the 5 minutes it took to cool to 52°C. Since the object is cooling, it cannot reach a lower temperature (27°C) before it reaches a higher temperature (52°C). The error stems from using the wrong logarithmic identity in line 5: t = ln(12)/k, not ln(12^(1/k)).gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution claims the time to reach 27° is approximately 17.92 minutes, but this is physically impossible because the object is already at 52° at t=5 minutes and cools towards 22°. It must take longer than 5 minutes to cool further to 27°. The correct time is approximately 22.92 minutes (5 * log(12)/log(2)). The error likely stems from a calculation mistake or misinterpretation of the logarithmic expression in the final step.
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 2026-10-10 with SymPy 1.14.0.