Newton's law of cooling
Problem 6.358 · hard
An object at 70° is placed in a room at 20°. After 20 minutes it has cooled to 60°. Using Newton's law of cooling, find its temperature after 60 minutes and when it reaches 25°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).Reviewed
- \[ \frac{d}{d t} \left(20 + 50 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}}\right) = - \frac{5 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} \ln{\left(\frac{5}{4} \right)}}{2} \]The model satisfies the cooling law.✓ Proved
- \[ 60 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{228}{5} \]T(60).✓ Proved
- \[ \frac{20 \ln{\left(10 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(10^{\frac{20}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(60) = \frac{228}{5} \approx 45.60^\circ,\quad t = \ln{\left(10^{\frac{20}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 206.38\text{ min} \)
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 identifies the model, determines the cooling constant k, and computes the required values. The final expression for time is algebraically equivalent to the standard form, though written in a non-standard way involving log(10).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the model, determines the cooling constant k, and computes the required values. The final expression for time is algebraically equivalent to the standard form, though written in a non-standard way involving log(10).gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly state the values of the constants (T_a, T_0, k) derived from the problem statement, making the subsequent equations opaque and difficult to verify. Furthermore, the final expression for time t is unnecessarily complex and not simplified to the standard form t = (20 * ln(10)) / ln(5/4), which obscures the calculation.
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.