Heat power needed to partially evaporate water from a water container

AI Thread Summary
To determine the power needed to evaporate 1 kg of water from a container with 3 kg of water, the heat energy required must be calculated using the formula Q = mcΔT + mL. The initial temperature of the water is 20 degrees Celsius, and the evaporation process requires accounting for both the temperature increase and the latent heat of vaporization. The challenge arises from the presence of the additional 2 kg of water, which complicates the temperature dynamics and energy distribution. It's essential to clarify whether the temperature is uniform throughout the container or if there are temperature variations in different parts of the water. Understanding these factors will help in accurately calculating the necessary power rating for the hot plate.
ezadam
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Homework Statement



So in the context of a practical project, I am tasked with selecting an appropriate hot plate with a specific power rating able to evaporate 1 kg of water in an hour, from a container that initially contains 3 kg of water.

Homework Equations



- P = Q/Δt

- Q = mcwaterΔT with c= 4.18 J/(g.K)

- Q = mL with L=2260 J/g

- Electricity-to-heat energy conversion efficiency: assumed to be 100% to simplify the problem. So the power from the hot plate goes directly to the water in question.

- Water in the container is initially at 20 degrees Celsius.

- Reminder: the container doesn't contain only the 1 kg of water that evaporates, it also has a total of 3 kg of water.

- NO INFO whatsoever about the container's dimensions and constitution or the hot stove itself. The container is exposed to ambient air.

Here is an illustrative sketch for the problem:

6h2pn7.jpg


The Attempt at a Solution



I first went like this:

P = Electrical energy/Δt = heat energy/Δt = heat energy/(3600 seconds)

Now my only problem is how to determine that heat energy. It would have been much easier if the container contained only the water to be evaporated, that is the 1 kg. In that case, it would have been:

Heat energy = mcwaterΔT + mL = 1000*4.18*(100-20) + 1000*2260
= some given number

Now I have no clue about what to do in the actual case, which is trying to evaporate 1kg of water from a total of 3 kg of water. Do I need extra equations in order to calculate that, or did I just misunderstand the concept ? It's been a while since my last Physics class ...
 
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ezadam said:
Now I have no clue about what to do in the actual case, which is trying to evaporate 1kg of water from a total of 3 kg of water. Do I need extra equations in order to calculate that, or did I just misunderstand the concept ? It's been a while since my last Physics class ...

Can separate parts of the liquid water in the container exist at different temperatures, or will the temperature be essentially uniform throughout the container?
 
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