Thermal Properties of Matter

Physics · Class 11

Simulation · Physics · Class 11

Newton's law of cooling

From the lesson Newton's law of cooling in Thermal Properties of Matter. Change the values and watch what happens.

Newton's law of coolingPhysics · Class 11

The idea behind it

NCERT §10.10

  • 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.