Newton's law of cooling
Problem 6.356 · hard
An object at 218° is placed in a room at 68°. After 10 minutes it has cooled to \frac{361}{2}°. Using Newton's law of cooling, find its temperature after 45 minutes and when it reaches 83°.
- T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
- \[ \frac{d}{d t} \left(68 + 150 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{10}}\right) = - 15 e^{- \frac{t \ln{\left(\frac{4}{3} \right)}}{10}} \ln{\left(\frac{4}{3} \right)} \]The model satisfies the cooling law.✓ Proved
- \[ \frac{361}{2} \]k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(4/3)/10 matches the reading at t = 10.✓ Proved
- \[ \frac{6075 \sqrt{3}}{256} + 68 \]T(45).✓ Proved
- \[ \frac{10 \ln{\left(10 \right)}}{\ln{\left(\frac{4}{3} \right)}} = \ln{\left(10^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \]Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved
Answer \( T(45) = \frac{6075 \sqrt{3}}{256} + 68 \approx 109.10^\circ,\quad t = \ln{\left(10^{\frac{10}{\ln{\left(\frac{4}{3} \right)}}} \right)} \approx 80.04\text{ min} \)
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 calculate the time required to reach 83°, instead providing a formula for a target temperature of 78° (which yields ~80.04 min). The problem explicitly asks for the time when the temperature reaches 83°.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to calculate the time required to reach 83°, instead providing a formula for a target temperature of 78° (which yields ~80.04 min). The problem explicitly asks for the time when the temperature reaches 83°.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution calculates the time to reach 83° as approximately 80 minutes, but the object is already at 180.5° at t=0 and 180.5° > 83°. Since the object is cooling towards 68°, it must pass through 83° much earlier than 80 minutes (specifically, around 10-20 minutes). The formula used in line 5 appears to have swapped the numerator and denominator or misidentified the target temperature relative to the ambient, leading to a physically impossible result where the object heats up or the time is grossly miscalculated. Specifically, T(45) is calculated as ~109°, which is correct, but the time to reach 83° should be less than 45 minutes. The provided answer t ≈ 80 min implies the object is still above 83° at t=80, which contradicts the cooling trend if it passed 109° at t=45. Wait, if T(45) ≈ 109, and it cools to 68, it will reach 83 at some t > 45. Let's re-evaluate. T(0)=180.5, T(10)=180.5, wait. 361/2 = 180.5. T_a = 68. T_0 - T_a = 112.5. T(10) - T_a = 112.5. This implies k=0? No. 361/2 = 180.5. 180.5 - 68 = 112.5. Initial temp 218. 218 - 68 = 150. So (T(10)-Ta)/(T0-Ta) = 112.5/150 = 0.75 = 3/4. So e^(-10k) = 3/4. k = -ln(3/4)/10 = ln(4/3)/10. This matches line 3. T(45) = 68 + 150 * e^(-45k) = 68 + 150 * (3/4)^(4.5). (3/4)^4.5 = (3/4)^4 * sqrt(3/4) = (81/256) * (sqrt(3)/2) = 81*sqrt(3)/512. 150 * 81*sqrt(3)/512 = 12150*sqrt(3)/512 = 6075*sqrt(3)/256. This matches line 4. Value is approx 109.1. Correct. Now for t when T(t)=83. 83 - 68 = 15. 15/150 = 1/10. e^(-kt) = 1/10. -kt = ln(1/10) = -ln(10). t = ln(10)/k = 10*ln(10)/ln(4/3). ln(10) ≈ 2.302. ln(4/3) ≈ 0.2877. t ≈ 10 * 2.302 / 0.2877 ≈ 80.04. The calculation is actually correct. My initial intuition was wrong because 109 to 83 is a small drop compared to the total range, but the cooling slows down. Let's check T(80). e^(-80k) = (3/4)^8. (0.75)^8 ≈ 0.1001. 150 * 0.1001 ≈ 15.01. 68 + 15.01 = 83.01. The math is correct. Why did I think it was wrong? I misjudged the rate. The solution is 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 2026-10-10 with SymPy 1.14.0.