Newton's law of cooling
Problem 6.279 · hard
An object at 40° is placed in a room at 0°. After 20 minutes it has cooled to 20°. Using Newton's law of cooling, find its temperature after 40 minutes and when it reaches 15°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} 40 e^{- \frac{t \ln{\left(2 \right)}}{20}} = - 2 e^{- \frac{t \ln{\left(2 \right)}}{20}} \ln{\left(2 \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 20 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(2)/20 matches the reading at t = 20.✓ Proved
- \[ 10 \]T(40).✓ Proved
- \[ \frac{20 \ln{\left(\frac{8}{3} \right)}}{\ln{\left(2 \right)}} = - \frac{20 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} + 60 \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(40) = 10 \approx 10.00^\circ,\quad t = - \frac{20 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} + 60 \approx 28.30\text{ min} \)
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 15° as approximately 28.3 minutes, which is before the object reached 20° at 20 minutes. This is physically impossible for a cooling object, indicating a sign error in the algebraic manipulation of the logarithmic terms (specifically, ln(8/3) was incorrectly equated to -ln(3) + 3ln(2) or similar, leading to a negative time offset relative to t=20).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution calculates the time to reach 15° as approximately 28.3 minutes, which is before the object reached 20° at 20 minutes. This is physically impossible for a cooling object, indicating a sign error in the algebraic manipulation of the logarithmic terms (specifically, ln(8/3) was incorrectly equated to -ln(3) + 3ln(2) or similar, leading to a negative time offset relative to t=20).qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly calculates the time to reach 15°. The formula used in line 5 corresponds to solving for a target temperature of 10° (sincegpt-oss:20b: inconclusive 2026-10-06 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The derivative in step 2 is incorrect (missing the factor 40) and the model is not explicitly stated as \(T(t)=40e^{-kt}\) with ambient 0. The subsequent
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 2026-10-06 with SymPy 1.14.0.