Heat power needed to partially evaporate water from a water container

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SUMMARY

The discussion focuses on calculating the power rating required for a hot plate to evaporate 1 kg of water from a container with 3 kg of water at an initial temperature of 20 degrees Celsius. The relevant equations include P = Q/Δt, Q = mcΔT, and Q = mL, with the specific heat capacity of water being 4.18 J/(g·K) and the latent heat of vaporization at 2260 J/g. The participant seeks clarification on whether additional equations are necessary to account for the total mass of water and the uniformity of temperature within the container.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the specific heat capacity and latent heat of vaporization.
  • Basic knowledge of power calculations in physics.
  • Ability to manipulate and apply equations related to energy and temperature changes.
NEXT STEPS
  • Research the concept of heat transfer in multi-phase systems.
  • Learn about the effects of container material and dimensions on heat transfer efficiency.
  • Explore advanced thermodynamic equations for systems with multiple phases.
  • Investigate the impact of ambient temperature on evaporation rates.
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Students in physics or engineering, educators teaching thermodynamics, and anyone involved in practical applications of heat transfer and evaporation processes.

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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