Newton's law of cooling
Problem 6.231 · hard
An object at 72° is placed in a room at 22°. After 15 minutes it has cooled to 62°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 32°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(22 + 50 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{15}}\right) = - \frac{10 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{15}} \ln{\left(\frac{5}{4} \right)}}{3} \]The model satisfies the cooling law.✓ Proved
- \[ 62 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/15 matches the reading at t = 15.✓ Proved
- \[ \frac{238}{5} \]T(45).✓ Proved
- \[ \frac{15 \ln{\left(5 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(5^{\frac{15}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = \frac{238}{5} \approx 47.60^\circ,\quad t = \ln{\left(5^{\frac{15}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 108.19\text{ min} \)
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 derived value of k or the specific function T(t) used in steps 4 and 5, making the algebraic jumps opaque. Furthermore, the final expression for t is unnecessarily complex and obscures the standard logarithmic form, which is pedagogically poor.
Every verdict on record (4)
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 in steps 4 and 5, making the algebraic jumps opaque. Furthermore, the final expression for t is unnecessarily complex and obscures the standard logarithmic form, which is pedagogically poor.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Newton's Law of Cooling, derives the cooling constant k accurately from the given data, and computes the requested values for T(45) and the time to reach 32°. The algebraic manipulations and final numerical approximations are correct.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.