Newton's law of cooling
Problem 6.399 · hard
An object at 60° is placed in a room at 0°. After 20 minutes it has cooled to 40°. Using Newton's law of cooling, find its temperature after 40 minutes and when it reaches 5°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} 60 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{20}} = - 3 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{20}} \ln{\left(\frac{3}{2} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 40 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(3/2)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{80}{3} \]T(40).✓ Proved
- \[ \frac{20 \ln{\left(12 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(12^{\frac{20}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(40) = \frac{80}{3} \approx 26.67^\circ,\quad t = \ln{\left(12^{\frac{20}{\ln{\left(\frac{3}{2} \right)}}} \right)} \approx 122.57\text{ min} \)
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 fails to explicitly state the values of T_a, T_0, and the derived constant k, making the subsequent equations opaque and difficult to verify without reverse-engineering. Furthermore, the final expression for t is algebraically correct but presented in a confusing, non-simplified form (log(12^(...))) rather than the standard t = (20 * ln(12)) / ln(1.5), which obscures the calculation logic.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to explicitly state the values of T_a, T_0, and the derived constant k, making the subsequent equations opaque and difficult to verify without reverse-engineering. Furthermore, the final expression for t is algebraically correct but presented in a confusing, non-simplified form (log(12^(...))) rather than the standard t = (20 * ln(12)) / ln(1.5), which obscures the calculation logic.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to explicitly state the value of the ambient temperature T_a (0°) in the setup, which is critical for the model. Furthermore, the final expression for the time to reach 5° is mathematically incorrect; it simplifies to t = 20 * ln(12) / ln(1.5), but the provided answer contains a nested logarithm in the exponent that does not simplify correctly to this value, and the derivation in step 5 is incomplete.
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.