What Happens to Water Temperature and State After Heating for 25 Minutes?

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The discussion focuses on calculating the final temperature and state of water heated by a 500 J/s heater over 25 minutes. The final state of the water is determined to be a combination of liquid and vapor, with a final temperature of 100°C. Participants emphasize the importance of understanding energy transfer, specifically using the formula Q = mcΔT to calculate the heat added. They guide the user in determining the total energy supplied by the heater and the energy required to raise the temperature and vaporize the water. The conversation highlights the need for a systematic approach to solving the problem based on the principles of thermodynamics.
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A heater so P=500Js^-1,heats up a tank of water of 200ml from 25°C for 25 minutes. What is the final form(solid ,liquid or gas)and the final temperature of the water?
The answer:
Final form : Liquid+vapour
Final temperature: 100°C

I have no idea how to do this!
Plz help!
 
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Welcome to PhysicsForums, Lennonlim123.

Start by understanding the physical situation. Even without looking at the answer, do you see that you are heating water, maybe until some or all of it evaporates?

Do you know how to figure out how much water heats from adding a given amount of energy?
 
I only know Q=mctheta
But how to apply it here?
Wait, I forgot.
The latent heat of vapourise is 2x10^6
The heat capacity of water is 4200
Heat capacity of vapour is 2100
 
Hi Lennonlim123. Welcome to PF!

Generally, you have to follow the rules and tell us what you have tried first. I will assume you don't know where to start. To get a start, answer these questions:

How much energy does the heater add to the water in 25 min.? (hint: you are given the amount of energy is added to the water each second).
How much energy does it take to raise a kg of water one degree C.? How much energy does it take to raise .2kg one degree C?
How much energy does it take to raise .2 kg of water to the boiling point?
How much energy does it take to vaporize one kg of water at that boiling point? How much for .2 kg of water?

(Note: assume the tank is at atmospheric pressure).

AM
 
Lennonlim123 said:
I only know Q=mctheta
I usually see that written as Q=mc∆T
It says that the the amount of heat (Q) added to a body equals the product of the mass (m), the heat capacity (c) and the change of temperature (∆T)

That let's you see where Andrew Mason's first leading question comes in:
How much energy does the heater add to the water in 25 min.? (hint: you are given the amount of energy is added to the water each second).

The answer to that will give you the value to substitute for Q (assuming all the heat goes into the water and stays there until the water boils).
 
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