Newton's law of cooling
Problem 6.240 · hard
An object at 150° is placed in a room at 70°. After 15 minutes it has cooled to 130°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 80°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(70 + 80 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{15}}\right) = - \frac{16 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{15}} \ln{\left(\frac{4}{3} \right)}}{3} \]The model satisfies the cooling law.✓ Proved
- \[ 130 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/15 matches the reading at t = 15.✓ Proved
- \[ \frac{415}{4} \]T(45).✓ Proved
- \[ \frac{15 \ln{\left(8 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(2^{\frac{45}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = \frac{415}{4} \approx 103.75^\circ,\quad t = \ln{\left(2^{\frac{45}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 108.42\text{ min} \)
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: fail (misleading) — The final sentence incorrectly equates the time to reach 80° with an unrelated logarithmic expression involving 45; the correct time is t = 15·ln(8)/ln(4/3).qwen3.6:27b-mlx: pass — The solution correctly applies Newton's Law of Cooling, derives the constant k accurately, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically complex.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Newton's Law of Cooling, derives the constant k accurately, and computes the requested values. The final expression for time is algebraically equivalent to the standard form, though stylistically complex.gpt-oss:20b: fail (misleading) 2026-10-04 — The final sentence incorrectly equates the time to reach 80° with an unrelated logarithmic expression involving 45; the correct time is t = 15·ln(8)/ln(4/3).qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to explicitly state the derived value of k or the specific function T(t) used for the calculations, making the transition from the general law to the specific numerical answers opaque. Furthermore, the expression for t in line 5 is unnecessarily complex and obscures the standard form t = ln((T0-Ta)/(T-Ta))/k, potentially confusing the student about how to solve for time.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.