Estimate the power requirement of the heater during this process. You may assume the specific heat of water is constant at

Like the flow calorimeter of a particular single-cup coffee maker uses an electric heating element to heat a steady flow of water from 22°C to 88°C. It heats 8 fluid ounces of water (with a mass of 237 g) in 60 s.



Flow calorimeter

What will be an ideal response?


The relevant steady-state energy balance is

?H = Q + W

Whether we regard the energy input as heat or work depends on where we draw the system boundary. If we draw the system boundary to only contain the fluid, then W = 0, and ?H = Q, where Q is the flow of heat from the heating element to the water. The total change in enthalpy of the 237 g of water is

= 65384 J = 65.4 kJ. This is also the total heat transferred to the water in a period of 60 s. The rate of heat transfer is therefore 65.4 kJ/60 s = 1.09 kJ/s = 1.09 kW. We can reasonably assume that the heating element fully converts electrical energy to heat, so at steady state, this would also be the power requirement for the heater. The actual power requirement might be slightly higher, because of heat losses to the surroundings (not all of the heat generated in the heating element actually enters the water). If the coffee maker is plugged into a regular U.S. 110 volt outlet, then it will draw at least 10 amps of current during the heating cycle. One wouldn’t want to run this along with much of anything else on a regular 15 amp household circuit.

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