Newton's law of cooling
Problem 6.272 · hard
An object at 150° is placed in a room at 0°. After 20 minutes it has cooled to \frac{225}{2}°. Using Newton's law of cooling, find its temperature after 40 minutes and when it reaches 10°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} 150 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{20}} = - \frac{15 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{20}} \ln{\left(\frac{4}{3} \right)}}{2} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{225}{2} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/20 matches the reading at t = 20.✓ Proved
- \[ \frac{675}{8} \]T(40).✓ Proved
- \[ \frac{20 \ln{\left(15 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(15^{\frac{20}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(40) = \frac{675}{8} \approx 84.38^\circ,\quad t = \ln{\left(15^{\frac{20}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 188.27\text{ min} \)
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 define the temperature function T(t) explicitly, making the subsequent algebraic checks (lines 2-5) disconnected from the problem statement. Furthermore, the final answer for the time to reach 10° is presented in a non-standard, overly complex logarithmic form rather than the simplified value derived in line 5.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution fails to define the temperature function T(t) explicitly, making the subsequent algebraic checks (lines 2-5) disconnected from the problem statement. Furthermore, the final answer for the time to reach 10° is presented in a non-standard, overly complex logarithmic form rather than the simplified value derived in line 5.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution calculates the time to reach 10° as approximately 188 minutes, which is less than the time it took to cool to 112.5° (20 minutes). This is physically impossible for a cooling object, indicating a sign error in the logarithmic calculation for t (the ratio inside the log should be inverted or the sign of k handled correctly to yield a time > 20).
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.