Question

200g of water at 30°C is put in a 1000 W (J/s) microwave oven. For simplicity assume that all the energy produced by the microwave goes into the water. The microwave runs for 2 minutes. Constants for H2O: TMP=0°C, TBP=100°C, ΔHmelt = 333.5 kJ/kg, ΔHvap = 2257 kJ/kg, cliquid=4.18 kJ/kgK, cgas=1.87 kJ/kgK. a) Calculate the total amount of energy transferred to the water. Show your work. b) Draw the Three-Phase Diagram (temperature versus energy added) and label the initial state of this process. Label the phase(s) of each section on the diagram and write down the corresponding energy equations in terms of relevant variables. This part is meant to help guide you through part c). c) At the end of the process, what is the final temperature and phase of H2O? If the final state is mixed phase, calculate the mass of each phase. You might find it helpful to draw energyinteraction diagram(s). Show your work.

Answer 1

(a)

The total energy produced by the microwave in 2 minutes i.e. 120 seconds is

Since all the energy produced by microwave is assumed to be transferred to the water, the total value of energy transferred to the warer is   .

(b)

Energy added to water as a function of increase in temperature, while in liquid phase,

Therefore the energy used in reaching   , starting from , is 836x70 = 58.52 k Joules .

The energy required to convert all of the 200 grams of water to vapour at the boiling point is

So, beacuse the total energy transferred from microwave to water is , not all of the water at the boiling point will convert to vapour.

The energy available to, partially, convert the water to vapor at the boiling point is

(c)

At the end, the 200 grams of is in mixed phase, water+vapour, at .

As the amount of energy available to convert water to vapour is 61.48 k Joules, the mass of   that becomes vapour is

So, we have approximately 27 grams of vapour and 173 grams of liquid water in the end.