Newton's law of cooling
Problem 6.357 · hard
An object at 82° is placed in a room at 22°. After 20 minutes it has cooled to 70°. Using Newton's law of cooling, find its temperature after 40 minutes and when it reaches 37°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(22 + 60 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}}\right) = - 3 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} \ln{\left(\frac{5}{4} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 70 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{302}{5} \]T(40).✓ Proved
- \[ \frac{20 \ln{\left(4 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(2^{\frac{40}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(40) = \frac{302}{5} \approx 60.40^\circ,\quad t = \ln{\left(2^{\frac{40}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 124.25\text{ min} \)
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 (error) — The calculation of T(40) is incorrect: using k=ln(5/4)/20 gives T(40)=22+60*(4/5)^2=41.2°, not 60.4°. The subsequent time to reach 37° is correct, but the earlier temperature value is wrong.qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly state the derived value of the cooling constant k or the specific temperature function T(t) before evaluating it. While the equations are algebraically correct, the logical flow skips the crucial step of defining the model parameters (k) and the function T(t) based on the given data, making the subsequent evaluations (lines 4 and 5) appear as ungrounded assertions rather than derived results.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-10 — The calculation of T(40) is incorrect: using k=ln(5/4)/20 gives T(40)=22+60*(4/5)^2=41.2°, not 60.4°. The subsequent time to reach 37° is correct, but the earlier temperature value is wrong.qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly state the derived value of the cooling constant k or the specific temperature function T(t) before evaluating it. While the equations are algebraically correct, the logical flow skips the crucial step of defining the model parameters (k) and the function T(t) based on the given data, making the subsequent evaluations (lines 4 and 5) appear as ungrounded assertions rather than derived results.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly state the derived value of the cooling constant k or the specific function T(t) used for the calculations, making the transition from the general law to the specific numerical answers opaque. Additionally, the final expression for time t is unnecessarily complex and obscures the standard linear form t = ln(ratio)/k.
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.