Newton's law of cooling.

A hot pan's T(t) decays exponentially toward room temperature; the ln(T − Ts) plot is straight and the average-temperature shortcut is checked.

With 94→86 °C in 2 min at 20 °C room, predict and verify the time for 71→69 °C (42 s).

Using the starting temperature instead of the average over the interval.

• Hot milk left on a table cools until it reaches room temperature. Measuring water heated about 40 °C above the room every minute shows the cooling is fast at first and slows as the gap to room temperature shrinks.
• Newton's law of cooling: for small temperature differences, the rate at which a body loses heat is proportional to the difference between its temperature T₂ and that of its surroundings T₁: −dQ/dt = k(T₂ − T₁).
• k depends on the area and nature of the surface. With dQ = ms dT₂ for a body of mass m and specific heat s: dT₂/dt = −K(T₂ − T₁), where K = k/(ms).
• Integrating: ln(T₂ − T₁) = −Kt + c, or T₂ = T₁ + C′e^(−Kt). The excess temperature decays exponentially with time.
• So a graph of ln(T₂ − T₁) against t is a straight line with negative slope. This is checked with a copper calorimeter of hot water inside a double-walled vessel whose walls hold water at a steady T₁.
• The law covers the combined loss by conduction, convection and radiation when the temperature difference is small: a radiator warming a room, heat leaking through a wall, a cup of tea cooling.
• Worked example: food in a pan cools from 94 °C to 86 °C in 2 min in a room at 20 °C. Average 90 °C, excess 70 °C: 8/2 = 70K, so K = 4/70 per minute.
• Cooling from 71 °C to 69 °C: average 70 °C, excess 50 °C, so 2/t = 50K = 50 × 4/70, which gives t = 0.7 min = 42 s.
• A shortcut for such problems: (T_a − T_b)/t = K[(T_a + T_b)/2 − T_s], which uses the average temperature over the interval.

Teaching this idea? Put the simulation on your own page or LMS with this code.