Newton's law of cooling
Problem 6.237 · hard
An object at 80° is placed in a room at 0°. After 20 minutes it has cooled to 64°. 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
- \[ \frac{d}{d t} 80 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} = - 4 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} \ln{\left(\frac{5}{4} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 64 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{128 \sqrt{5}}{5} \]T(30).✓ Proved
- \[ \frac{20 \ln{\left(8 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(2^{\frac{60}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{128 \sqrt{5}}{5} \approx 57.24^\circ,\quad t = \ln{\left(2^{\frac{60}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 186.38\text{ min} \)
Lines: 4 proved, 1 reviewed. 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 | 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: passqwen3.6:27b-mlx: pass — The solution correctly applies Newton's Law of Cooling, derives the constant k, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically complex.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Newton's Law of Cooling, derives the constant k, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically complex.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to explicitly state the derived function T(t) or the value of k before using them in subsequent steps, making the logical flow disjointed. Furthermore, the final expression for t is unnecessarily complex and obscures the standard logarithmic form, though algebraically correct.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.