Newton's law of cooling
Problem 6.317 · hard
An object at 110° is placed in a room at 70°. After 20 minutes it has cooled to 102°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 80°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(70 + 40 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}}\right) = - 2 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} \ln{\left(\frac{5}{4} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 102 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{64 \sqrt{5}}{5} + 70 \]T(30).✓ Proved
- \[ \frac{20 \ln{\left(4 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(2^{\frac{40}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{64 \sqrt{5}}{5} + 70 \approx 98.62^\circ,\quad t = \ln{\left(2^{\frac{40}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 124.25\text{ min} \)
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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Sentence 5 incorrectly derives the time to reach 80°. The correct expression is \(t=20\ln 4/\ln(5/4)\); the given logarithmic manipulation is algebraicallqwen3.6:27b-mlx: fail (error) — The solution fails to explicitly state the derived value of the cooling constant k or the specific function T(t) used for the calculations, making the transition from the general law to the specific numerical answers opaque. Additionally, the expression for the time to reach 80° is algebraically correct but presented in a confusingly complex form that obscures the standard solution t = (1/k)ln(40/10).
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly state the derived value of the cooling constant k or the specific function T(t) used for the calculations, making the transition from the general law to the specific numerical answers opaque. Additionally, the expression for the time to reach 80° is algebraically correct but presented in a confusingly complex form that obscures the standard solution t = (1/k)ln(40/10).gpt-oss:20b: inconclusive 2026-10-08 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Sentence 5 incorrectly derives the time to reach 80°. The correct expression is \(t=20\ln 4/\ln(5/4)\); the given logarithmic manipulation is algebraicallqwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly state the value of k or the specific function T(t) used to derive the answers, making the jump from the general law to the specific numerical results opaque. Furthermore, the expression for the time to reach 80° is unnecessarily complex and obscures the standard logarithmic form, though algebraically equivalent.gpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.