Newton's law of cooling
Problem 6.280 · hard
An object at 80° is placed in a room at 20°. After 15 minutes it has cooled to 65°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 30°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(20 + 60 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{15}}\right) = - 4 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{15}} \ln{\left(\frac{4}{3} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 65 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/15 matches the reading at t = 15.✓ Proved
- \[ \frac{725}{16} \]T(45).✓ Proved
- \[ \frac{15 \ln{\left(6 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(6^{\frac{15}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = \frac{725}{16} \approx 45.31^\circ,\quad t = \ln{\left(6^{\frac{15}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 93.42\text{ min} \)
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 (misleading) — The solution fails to explicitly state the values of T_a, T_0, and the derived constant k, making the transition from the general formula to the specific numerical checks opaque. Additionally, the final expression for time is unnecessarily complex and obscures the standard form t = (1/k)ln((T_0 - T_a)/(T - T_a)).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (misleading) 2026-10-06 — The solution fails to explicitly state the values of T_a, T_0, and the derived constant k, making the transition from the general formula to the specific numerical checks opaque. Additionally, the final expression for time is unnecessarily complex and obscures the standard form t = (1/k)ln((T_0 - T_a)/(T - T_a)).qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies Newton's Law of Cooling, derives the constant k, and computes the requested values accurately. The final expression for time is algebraically equivalent to the standard form.gpt-oss:20b: pass 2026-10-06
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.