Newton's law of cooling
Problem 6.355 · hard
An object at 75° is placed in a room at 0°. After 10 minutes it has cooled to \frac{225}{4}°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 15°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} 75 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{10}} = - \frac{15 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{10}} \ln{\left(\frac{4}{3} \right)}}{2} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{225}{4} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/10 matches the reading at t = 10.✓ Proved
- \[ \frac{6075 \sqrt{3}}{512} \]T(45).✓ Proved
- \[ \frac{10 \ln{\left(5 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(5^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = \frac{6075 \sqrt{3}}{512} \approx 20.55^\circ,\quad t = \ln{\left(5^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 55.95\text{ min} \)
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 state the derived function T(t) with the calculated constant k, jumping from the general form and parameter verification directly to specific evaluations. While the algebraic checks are correct, the logical flow is broken because the specific model T(t) = 75 * exp(-t * ln(4/3)/10) is used in the equation checks but never formally established as the result of the setup steps.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly state the derived function T(t) with the calculated constant k, jumping from the general form and parameter verification directly to specific evaluations. While the algebraic checks are correct, the logical flow is broken because the specific model T(t) = 75 * exp(-t * ln(4/3)/10) is used in the equation checks but never formally established as the result of the setup steps.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly state the temperature function T(t) with the calculated constant k, jumping directly to unchecked algebraic identities. Furthermore, the final answer for time t is presented in a mathematically valid but pedagogically confusing and non-standard form (log(5^(...))) rather than the simplified t = 10*ln(5)/ln(4/3), which obscures the result.
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 2026-10-10 with SymPy 1.14.0.