Newton's law of cooling
Problem 6.232 · hard
An object at 110° is placed in a room at 70°. After 5 minutes it has cooled to 100°. Using Newton's law of cooling, find its temperature after 15 minutes and when it reaches 80°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).Reviewed
- \[ \frac{d}{d t} \left(70 + 40 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{5}}\right) = - 8 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{5}} \ln{\left(\frac{4}{3} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 100 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/5 matches the reading at t = 5.✓ Proved
- \[ \frac{695}{8} \]T(15).✓ Proved
- \[ \frac{5 \ln{\left(4 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(2^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(15) = \frac{695}{8} \approx 86.88^\circ,\quad t = \ln{\left(2^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 24.09\text{ min} \)
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 accurately. The algebraic forms in the final answer are equivalent to the standard logarithmic solutions.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Newton's Law of Cooling, derives the decay constant k from the given data, and computes the requested values accurately. The algebraic forms in the final answer are equivalent to the standard logarithmic solutions.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Newton's Law of Cooling, derives the constant k accurately, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically complex.gpt-oss:20b: fail (error) 2026-10-04 — The solution’s third sentence is incorrect: it states “100 = 100” and claims to compute k, but does not actually use the data to solve for k. The model’s parameters are not properly derived from the given temperatures, so the subsequent calculations are not justified.
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 2026-10-04 with SymPy 1.14.0.