Cooking Thermodynamics

The change in internal energy is equal to the energy absorbed by the food

\[ \Delta U = Q \\ m \bar{c_p} \Delta T = \bar{\dot{Q}} \Delta t \]

Water temperature increases from \(25\) to \(100^\circ\text{C}\)

\[\Delta T=75^\circ\text{C}\]

\[ \bar{c_p}= 4.2 \text{ kJ/kg-K}\]

Heaters are about \(2000 \text{ W}\). Let's say \(10\%\) gets lost to the environment

\[ \bar{\dot{Q}} = 1.8 \text{ kW} \]

Substituting in the energy balance equation above, and making the time in minutes

\[ m \text{ [kg]} (4.2)(75) = (1.8) \Delta t \text{[min]} (60) \]

\[ \Delta t \text{[min]} \approx 3 m\text{ [kg]} \]