Quenching a steel bush: which cooling model to post on the shop floor
A foundry's quality lab quenches a small steel bush at 300 °C and leaves it in a room held at 30 °C. Newton's law of cooling applies: the rate of fall of temperature is proportional to the excess over room temperature. A trainee records 180 °C after exactly 10 minutes and 113 °C after 20 minutes. The process sheet allows the bush to be handled bare-handed only below 60 °C, and the shop floor wants a single handling time posted at the machine. A supervisor looks at the first reading only and argues that the fall is 12 °C per minute, so the bush will reach 60 °C at t = 20 min and room temperature at t = 22.5 min; he proposes posting 20 minutes and telling the stores clerk that the bush is “cold” by half past the half-hour. The trainee doubts both numbers but has only the readings, not the theory.
1. Set up and solve the differential equation for the temperature, and find the constant from the 10-minute reading. Then compute the time at which the bush reaches 60 °C.
Read the worked answer
Step 1 - model. Let T(t) be the temperature in °C and let the room be 30 °C. Newton's law gives
dT/dt = −k(T − 30), k > 0 (the minus sign because T is falling).
Step 2 - solve. Separate: dT/(T − 30) = −k dt ⇒ ln(T − 30) = −kt + C ⇒ T − 30 = A e^(−kt). With T(0) = 300, A = 270:
T(t) = 30 + 270 e^(−kt).
Step 3 - find k. At t = 10, T = 180: 150 = 270 e^(−10k) ⇒ e^(−10k) = 5/9. So k = (1/10)ln(9/5) ≈ 0.0588 min⁻¹. (Check with the second reading: T(20) = 30 + 270(5/9)² = 30 + 83.3 = 113.3 °C, matching the trainee's 113 °C.)
Step 4 - time to 60 °C. 60 − 30 = 30 = 270 e^(−kt) ⇒ e^(−kt) = 1/9 ⇒ t = ln 9 / k = 10 ln 9 / ln(9/5) ≈ 2.1972 / 0.0588 ≈ 37.4 minutes.
Post about 37 minutes as the earliest bare-hand time.
2. Evaluate the supervisor's linear model using the trainee's 20-minute reading and the behaviour of the solution as t grows. Which of his two numbers is defensible?
Read the worked answer
Step 1 - test the linear claim at t = 20. A constant 12 °C per minute predicts 300 − 12(20) = 60 °C at 20 minutes, but the recorded value is 113 °C. The linear model is out by 53 °C - it cools far too fast, because it ignores the fact that the driving difference (T − 30) shrinks as the bush cools.
Step 2 - test the 22.5-minute claim. The exact solution gives T − 30 = 270 e^(−kt) > 0 for every finite t, so T approaches 30 °C only asymptotically. Reaching room temperature in finite time is impossible under this model: setting T = 30 forces 270 e^(−kt) = 0, which has no solution. The claim that the bush is “cold” at 22.5 minutes is therefore not merely inaccurate, it is structurally impossible.
Step 3 - verdict. Neither of the supervisor's numbers survives. The 60 °C crossing is at t ≈ 37.4 min, not 20 min, and there is no finite time at which the bush equals room temperature. His error is using a chord between two points as if the curve were a straight line; Newton's law produces exponential decay, and it is the ratio of excess temperatures that falls at a constant rate, not the temperature itself.
3. The shop floor wants one number to post. Give a defensible handling time with a safety margin, and state the assumptions your number relies on.
Read the worked answer
Step 1 - start from the model answer. The computed crossing is t ≈ 37.4 min, so 37 minutes is already unsafe; rounding down would have the worker touch a bush still above 60 °C.
Step 2 - add a margin for the things that are not modelled. Recommend posting 45 minutes (≈ 37.4 min rounded up plus roughly 7 minutes of margin). Justification: the model assumes the room stays at 30 °C, that the bush is small enough for its temperature to be treated as uniform, and that k is constant over the whole range - none of these is guaranteed on a shop floor where doors open.
Step 3 - check the margin is affordable. T(45) = 30 + 270 e^(−0.0588 × 45) = 30 + 270 e^(−2.646) ≈ 30 + 270(0.0709) ≈ 49 °C, comfortably below 60 °C. So 45 minutes is safe without being absurdly conservative.
Step 4 - state the limits honestly. One bush, one room temperature, one pair of readings; k should be re-estimated if the ambient temperature or the bush size changes, and a real procedure would also verify with a thermometer rather than trusting the post alone.