Newton's law of cooling
Problem 6.312 · hard
An object at 100° is placed in a room at 25°. After 20 minutes it has cooled to \frac{325}{4}°. Using Newton's law of cooling, find its temperature after 60 minutes and when it reaches 40°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).Reviewed
- \[ \frac{d}{d t} \left(25 + 75 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{20}}\right) = - \frac{15 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{20}} \ln{\left(\frac{4}{3} \right)}}{4} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{325}{4} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{3625}{64} \]T(60).✓ Proved
- \[ \frac{20 \ln{\left(5 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(5^{\frac{20}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(60) = \frac{3625}{64} \approx 56.64^\circ,\quad t = \ln{\left(5^{\frac{20}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 111.89\text{ min} \)
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 decay constant k from the given data, and computes the requested values. The final expression for time t is algebraically equivalent to the standard form, though written in a redundant way.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies Newton's Law of Cooling, derives the decay constant k from the given data, and computes the requested values. The final expression for time t is algebraically equivalent to the standard form, though written in a redundant way.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly state the derived value of k or the specific function T(t) used for the calculations, making the subsequent equation checks opaque and disconnected from the problem statement. Furthermore, the final answer for time is presented in a redundant and non-standard form (log of a power) rather than the simplified expression derived in step 5.gpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.