Newton's law of cooling
Problem 6.233 · hard
An object at 145° is placed in a room at -5°. After 10 minutes it has cooled to 95°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 0°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(-5 + 150 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{10}}\right) = - 15 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{10}} \ln{\left(\frac{3}{2} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 95 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(3/2)/10 matches the reading at t = 10.✓ Proved
- \[ \frac{355}{9} \]T(30).✓ Proved
- \[ \frac{10 \ln{\left(30 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(30^{\frac{10}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{355}{9} \approx 39.44^\circ,\quad t = \ln{\left(30^{\frac{10}{\ln{\left(\frac{3}{2} \right)}}} \right)} \approx 83.88\text{ min} \)
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 fails to calculate the temperature at 30 minutes as requested, providing only the symbolic equality for T(30) without stating the final value. Additionally, the expression for the time to reach 0° is mathematically incorrect and does not simplify to the stated numerical answer.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to calculate the temperature at 30 minutes as requested, providing only the symbolic equality for T(30) without stating the final value. Additionally, the expression for the time to reach 0° is mathematically incorrect and does not simplify to the stated numerical answer.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to calculate the temperature at 30 minutes as requested, providing only the symbolic equality for T(30) without evaluating it or stating the final value. Additionally, the expression for the time to reach 0° is algebraically convoluted and potentially misleading, though numerically correct.gpt-oss:20b: pass 2026-10-04
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.