Newton's law of cooling
Problem 6.274 · hard
An object at 118° is placed in a room at 68°. After 10 minutes it has cooled to \frac{304}{3}°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 73°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).Reviewed
- \[ \frac{d}{d t} \left(68 + 50 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{10}}\right) = - 5 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{10}} \ln{\left(\frac{3}{2} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{304}{3} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(3/2)/10 matches the reading at t = 10.✓ Proved
- \[ \frac{2236}{27} \]T(30).✓ Proved
- \[ \frac{10 \ln{\left(10 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(10^{\frac{10}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{2236}{27} \approx 82.81^\circ,\quad t = \ln{\left(10^{\frac{10}{\ln{\left(\frac{3}{2} \right)}}} \right)} \approx 56.79\text{ min} \)
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, and computes the requested values. The final expression for time is algebraically equivalent to the standard logarithmic form.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies Newton's Law of Cooling, derives the decay constant k, and computes the requested values. The final expression for time is algebraically equivalent to the standard logarithmic form.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution fails to explicitly state the derived value of k or the specific temperature function T(t) before calculating T(30) and the time for 73°. While the intermediate equations are algebraically correct, the logical flow is broken because the model parameters are not formally established in the text, making the subsequent calculations appear unmotivated. Additionally, the final expression for time is unnecessarily complex and obscures the standard form t = (1/k)ln((T0-Ta)/(T-Ta)).
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.