Newton's law of cooling
Problem 6.275 · hard
An object at 45° is placed in a room at -5°. After 20 minutes it has cooled to 20°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 10°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(-5 + 50 e^{- \frac{t \ln{\left(2 \right)}}{20}}\right) = - \frac{5 e^{- \frac{t \ln{\left(2 \right)}}{20}} \ln{\left(2 \right)}}{2} \]The model satisfies the cooling law.✓ Proved
- \[ 20 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(2)/20 matches the reading at t = 20.✓ Proved
- \[ -5 + \frac{25 \cdot 2^{\frac{3}{4}}}{4} \]T(45).✓ Proved
- \[ \frac{20 \ln{\left(\frac{10}{3} \right)}}{\ln{\left(2 \right)}} = \ln{\left(\left(\frac{10}{3}\right)^{\frac{20}{\ln{\left(2 \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = -5 + \frac{25 \cdot 2^{\frac{3}{4}}}{4} \approx 5.51^\circ,\quad t = \ln{\left(\left(\frac{10}{3}\right)^{\frac{20}{\ln{\left(2 \right)}}} \right)} \approx 34.74\text{ min} \)
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 values of T_a, T_0, and the derived k in the sentences, making the jump to the specific function in line 2 unjustified without re-deriving it. More critically, the final answer for time t is presented as a complex logarithmic expression that simplifies to the correct value but is not clearly derived from the standard formula t = (1/k) * ln((T_0 - T_a)/(T - T_a)), leading to a confusing and non-standard final form that obscures the method.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution fails to explicitly state the values of T_a, T_0, and the derived k in the sentences, making the jump to the specific function in line 2 unjustified without re-deriving it. More critically, the final answer for time t is presented as a complex logarithmic expression that simplifies to the correct value but is not clearly derived from the standard formula t = (1/k) * ln((T_0 - T_a)/(T - T_a)), leading to a confusing and non-standard final form that obscures the method.gpt-oss:20b: fail (error) 2026-10-06 — The derivative in line 2 is incorrect: the coefficient should be 50, not –5, and the sign of the exponential term is wrong. This mis‑derivation undermines the subsequent steps.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution fails to state the final answers explicitly as required by the problem ('find its temperature... and when...'). It only provides intermediate algebraic identities and an incorrect symbolic form for the time calculation (using log base e inside a log base 2 exponent structure incorrectly, though the numerical value is coincidentally close, the formula presented is not the standard solution for t).
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.