Newton's law of cooling
Problem 6.236 · hard
An object at 97° is placed in a room at 22°. After 10 minutes it has cooled to \frac{313}{4}°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 32°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(22 + 75 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{10}}\right) = - \frac{15 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{10}} \ln{\left(\frac{4}{3} \right)}}{2} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{313}{4} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/10 matches the reading at t = 10.✓ Proved
- \[ \frac{3433}{64} \]T(30).✓ Proved
- \[ \frac{10 \ln{\left(\frac{15}{2} \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(\left(\frac{15}{2}\right)^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{3433}{64} \approx 53.64^\circ,\quad t = \ln{\left(\left(\frac{15}{2}\right)^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 70.04\text{ min} \)
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 explicitly define the variables T_a, T_0, and the derived constant k in the text, making the jump from the general formula to the specific equation in line 2 opaque. Furthermore, the final answer for time t is presented in a mathematically valid but unnecessarily complex and non-standard form (log of a power), whereas the simplified form t = 10 * ln(15/2) / ln(4/3) is standard and clearer.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to explicitly define the variables T_a, T_0, and the derived constant k in the text, making the jump from the general formula to the specific equation in line 2 opaque. Furthermore, the final answer for time t is presented in a mathematically valid but unnecessarily complex and non-standard form (log of a power), whereas the simplified form t = 10 * ln(15/2) / ln(4/3) is standard and clearer.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to explicitly state the derived model T(t) = 22 + 75e^(-kt) with the calculated k, which is necessary to justify the subsequent algebraic steps. Furthermore, the final expression for t is mathematically incorrect due to a logarithm base error: t = ln(A)/k simplifies to ln(A)/ln(B), not ln(A^(1/ln(B))). The provided answer t = ln((15/2)^(10/ln(4/3))) evaluates to approximately 24.5, not 70.04.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.