Newton's law of cooling
Problem 6.391 · hard
An object at 80° is placed in a room at 20°. After 10 minutes it has cooled to 60°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 25°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).Reviewed
- \[ \frac{d}{d t} \left(20 + 60 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{10}}\right) = - 6 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{10}} \ln{\left(\frac{3}{2} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 60 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(3/2)/10 matches the reading at t = 10.✓ Proved
- \[ \frac{320 \sqrt{6}}{81} + 20 \]T(45).✓ Proved
- \[ \frac{10 \ln{\left(12 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(12^{\frac{10}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = \frac{320 \sqrt{6}}{81} + 20 \approx 29.68^\circ,\quad t = \ln{\left(12^{\frac{10}{\ln{\left(\frac{3}{2} \right)}}} \right)} \approx 61.29\text{ min} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 cooling constant k, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though written in a slightly unusual exponential-logarithmic format.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly applies Newton's Law of Cooling, derives the cooling constant k, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though written in a slightly unusual exponential-logarithmic format.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly sets up Newton's Law of Cooling, determines the decay constant k, and derives the correct expressions for the temperature at t=45 and the time to reach 25°. The algebraic forms in the stated answer match the derived equations.
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 2026-10-12 with SymPy 1.14.0.