Newton's law of cooling
Problem 6.278 · hard
An object at 50° is placed in a room at 0°. After 5 minutes it has cooled to \frac{100}{3}°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 10°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} 50 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{5}} = - 10 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{5}} \ln{\left(\frac{3}{2} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{100}{3} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(3/2)/5 matches the reading at t = 5.✓ Proved
- \[ \frac{3200}{729} \]T(30).✓ Proved
- \[ \frac{5 \ln{\left(5 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(5^{\frac{5}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = \frac{3200}{729} \approx 4.39^\circ,\quad t = \ln{\left(5^{\frac{5}{\ln{\left(\frac{3}{2} \right)}}} \right)} \approx 19.85\text{ min} \)
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: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution calculates the time to reach 10° as approximately 19.85 minutes, which is less than the 5 minutes it took to cool to 33.3°. Since the o
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution calculates the time to reach 10° as approximately 19.85 minutes, which is less than the 5 minutes it took to cool to 33.3°. Since the ogpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies Newton's Law of Cooling, derives the correct cooling constant k, and computes the temperature at t=30 and the time to reach 10 degrees accurately. The algebraic manipulations and final numerical approximations are correct.
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.