How much water evaporates per minute from a copper pot on a stove?

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Homework Help Overview

The discussion revolves around the evaporation of water from a copper pot placed on a stove, with specific parameters including the pot's dimensions and temperatures of the stove and water. The problem involves understanding heat transfer and phase changes in the context of thermodynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the rate of heat transfer using given equations and seeks validation of their approach. Some participants question the applicability of the second equation due to the water being at its boiling point, prompting a discussion on phase changes and the nature of heat transfer during evaporation.

Discussion Status

Participants are actively engaging with the problem, providing insights on the correct equations to use during phase transitions. There is a recognition of the need to reconsider the original poster's approach, particularly regarding the implications of the water's temperature and the concept of latent heat.

Contextual Notes

Participants note the importance of understanding the phase change from liquid to gas and the associated energy considerations, highlighting potential confusion around temperature changes during evaporation.

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Homework Statement


a copper pot's radius is 12 cm, the thickness is 5 mm. It is placed on the stove for boiling, if the temperature of the stove is 115 oC while the water is 100 oC. How much of water that evaporates per minute

Homework Equations


ΔQ/Δt = (A)(k)(ΔT) / (L)
Q=mcΔT

The Attempt at a Solution



I 've calculated the rate of heat transferring in a minute with the first equation and I got the amount of the heat to calculate mass later with the second equation. Does my process correct?
 
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The method for calculating the heat per minute going in is correct. You need to rethink the second equation. The water is already at the boiling point so any heat that goes in it cannot raise its temperature. What does it do instead?
 
kuruman said:
The method for calculating the heat per minute going in is correct. You need to rethink the second equation. The water is already at the boiling point so any heat that goes in it cannot raise its temperature. What does it do instead?
it is changed into gas - steam? I know that when then heat reach the phase change, the temperature won't change anymore. How can I apply this with the second equation? Does the change in temperature still the same at 100oC because if there is no change in the temperature the whole equation will broke down (divided by 0)
 
You should not be looking at Q=mcΔT. What equation is applicable when you have a phase transition?
 
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In evaporation all energy absorbed is going to be latent and not sensible. Use this equation q=hfg [J/kg]. hfg represents the enthalpy of evaporation, this means, the specific energy required to go from one side of the dome(saturated liq) to the other side of the dome (saturated steam).
 

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