∫Calc Practice

Newton's law of cooling

Problem 6.351 · hard

An object at 145° is placed in a room at -5°. After 20 minutes it has cooled to 115°. Using Newton's law of cooling, find its temperature after 30 minutes and when it reaches 5°.
  1. T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
  2. \[ \frac{d}{d t} \left(-5 + 150 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}}\right) = - \frac{15 e^{- \frac{t \ln{\left(\frac{5}{4} \right)}}{20}} \ln{\left(\frac{5}{4} \right)}}{2} \]
    The model satisfies the cooling law.✓ Proved
  3. \[ 115 \]
    k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(5/4)/20 matches the reading at t = 20.✓ Proved
  4. \[ -5 + 48 \sqrt{5} \]
    T(30).✓ Proved
  5. \[ \frac{20 \ln{\left(15 \right)}}{\ln{\left(\frac{5}{4} \right)}} = \ln{\left(15^{\frac{20}{\ln{\left(\frac{5}{4} \right)}}} \right)} \]
    Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(30) = -5 + 48 \sqrt{5} \approx 102.33^\circ,\quad t = \ln{\left(15^{\frac{20}{\ln{\left(\frac{5}{4} \right)}}} \right)} \approx 242.72\text{ min} \)

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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0k fitted by a root-finder, then the cooling law integrated numerically

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly state the derived value of the cooling constant k or the specific temperature function T(t) used to calculate the final answers. While the algebraic checks in lines 2-5 are correct, the logical gap between determining k (line 3) and applying it to find T(30) and the time for 5° (lines 4-5) is not bridged by any sentence defining the model parameters, making the derivation incomplete.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-10
  • qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly state the derived value of the cooling constant k or the specific temperature function T(t) used to calculate the final answers. While the algebraic checks in lines 2-5 are correct, the logical gap between determining k (line 3) and applying it to find T(30) and the time for 5° (lines 4-5) is not bridged by any sentence defining the model parameters, making the derivation incomplete.
  • gpt-oss:20b: pass 2026-10-10
  • qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies Newton's Law of Cooling, derives the decay constant k, and computes the requested values. The algebraic forms in the stated answer match the intermediate steps.

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 2026-10-10 with SymPy 1.14.0.