Heat power needed to partially evaporate water from a water container

In summary, the task at hand is to select a hot plate with a specific power rating that can evaporate 1 kg of water in an hour from a container containing 3 kg of water. The given equations and assumptions are used to calculate the necessary heat energy and the temperature of the water in the container. There is uncertainty about whether additional equations are needed to account for the presence of 3 kg of water in the container and the potential variation in temperature throughout the container.
  • #1
ezadam
21
0

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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  • #2
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?
 

1. How do you calculate the heat power needed to partially evaporate water from a water container?

To calculate the heat power needed, you will need to know the amount of water to be evaporated, the specific heat capacity of water, and the latent heat of vaporization. The formula for heat power is Q = m * c * ΔT, where Q is the heat power in watts, m is the mass of water in kilograms, c is the specific heat capacity of water, and ΔT is the change in temperature. You will then need to multiply the heat power by the latent heat of vaporization to account for the energy needed to change the water from liquid to gas.

2. What is the specific heat capacity of water?

The specific heat capacity of water is the amount of heat energy required to raise the temperature of 1 kilogram of water by 1 degree Celsius. It is approximately 4.184 joules per gram per degree Celsius.

3. What is the latent heat of vaporization for water?

The latent heat of vaporization for water is the amount of heat energy required to change 1 kilogram of water from liquid to gas at its boiling point. It is approximately 2,260 joules per gram.

4. How does the amount of water affect the heat power needed for partial evaporation?

The amount of water directly affects the heat power needed for partial evaporation. The more water there is, the more heat power is required to raise its temperature and change it to gas. This is because it takes more energy to heat a larger amount of water and to change it to gas compared to a smaller amount.

5. Does the heat power needed for partial evaporation vary with the temperature of the water?

Yes, the heat power needed for partial evaporation does vary with the temperature of the water. This is because the specific heat capacity of water changes with temperature. As the temperature increases, the specific heat capacity decreases, meaning less heat power is needed to raise the temperature and change the water to gas.

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