Newton's law of cooling
Problem 6.277 · hard
An object at 105° is placed in a room at 25°. After 10 minutes it has cooled to 65°. Using Newton's law of cooling, find its temperature after 20 minutes and when it reaches 40°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(25 + 80 e^{- \frac{t \ln{\left(2 \right)}}{10}}\right) = - 8 e^{- \frac{t \ln{\left(2 \right)}}{10}} \ln{\left(2 \right)} \]The model satisfies the cooling law.✓ Proved
- \[ 65 \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(2)/10 matches the reading at t = 10.✓ Proved
- \[ 45 \]T(20).✓ Proved
- \[ \frac{10 \ln{\left(\frac{16}{3} \right)}}{\ln{\left(2 \right)}} = - \frac{10 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} + 40 \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(20) = 45 \approx 45.00^\circ,\quad t = - \frac{10 \ln{\left(3 \right)}}{\ln{\left(2 \right)}} + 40 \approx 24.15\text{ min} \)
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 (error) — The final step incorrectly solves for the time to reach 40°. The correct formula is t = (10/ln2)·ln(80/15) ≈ 24.15 min, not –10·ln3/ln2 + 40. The provided expression does not follow from the previous equations and would mislead a student.qwen3.6:27b-mlx: fail (error) — The solution contains a fatal algebraic error in step 5: it claims 10*log(16/3)/log(2) equals -10*log(3)/log(2) + 40, but the correct value for t when T=40 is 10*log(16/3)/log(2) ≈ 24.15, whereas the expression on the right evaluates to ≈ 4.15. The stated answer formula is incorrect.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The final step incorrectly solves for the time to reach 40°. The correct formula is t = (10/ln2)·ln(80/15) ≈ 24.15 min, not –10·ln3/ln2 + 40. The provided expression does not follow from the previous equations and would mislead a student.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution contains a fatal algebraic error in step 5: it claims 10*log(16/3)/log(2) equals -10*log(3)/log(2) + 40, but the correct value for t when T=40 is 10*log(16/3)/log(2) ≈ 24.15, whereas the expression on the right evaluates to ≈ 4.15. The stated answer formula is incorrect.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution contains a fatal algebraic error in the final step. The equation `10*log(16/3)/log(2) = -10*log(3)/log(2) + 40` is mathematically false (LHS ≈ 4.42, RHS ≈ 24.15). The correct time to reach 40° is `t = 10 * ln(16/3) / ln(2) ≈ 4.42` minutes, not 24.15 minutes. The stated answer formula `t = -10 ln(3)/ln(2) + 40` is nonsensical in this context.
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.