A typical 12v batter can deliver about 7.5 x 10^5 C

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Homework Help Overview

The problem involves calculating the maximum mass of water that can be converted into steam using the energy from a typical 12v battery, which can deliver approximately 7.5 x 10^5 C of charge. The subject area pertains to thermodynamics and electrical energy conversion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the formula E=qv to calculate energy in Joules and question how to apply this to the boiling of water. There is a focus on understanding the latent heat of vaporization and its relevance to the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on the necessary energy calculations and clarifying the steps needed to relate the battery's energy output to the boiling of water. There is an exploration of the energy required for phase change and how it connects to the available energy from the battery.

Contextual Notes

Participants note the need to look up the latent heat of vaporization for water at 100 degrees Celsius, indicating a reliance on external data for completing the calculations. There is also a recognition of the importance of careful reading of the problem statement.

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PROBLEM: A typical 12v battery can deliver about 7.5 x 10^5 C of charge before dying. This is not very much. To get a feel for this calculate the maximum number of kilograms of water (100 degrees celsius) that could be boiled into steam (100degrees celsius) using energy from this battery.

I think I'm supposed to use the formula, E= qv, but I don't know what to do after that.
 
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Using E=qv, your energy will be in units of Joules. How many Joules does it take to raise 1 kilogram of water 1 degree celsius?

Edit: Oops. After re-reading the question, I see that the intent is to bring liquid water from 100C to vapor at 100C. For this, look up Latent heat of vaporization of water.
 
Last edited:
Phrak said:
Using E=qv, your energy will be in units of Joules. How many Joules does it take to raise 1 kilogram of water 1 degree celsius?

Edit: Oops. After re-reading the question, I see that the intent is to bring liquid water from 100C to vapor at 100C. For this, look up Latent heat of vaporization of water.

Thanks for responding. So all I have to do is look up that value for the latent heat of vaporization and that would be my final answer...I don't have to use any formula to calculate anything? I looked up the value and for 100 degrees C, the latent heat is 2260 kj/kg...would that be all?
 
Read the question carefully, then:-

So what you've got, is that you need 2260kJ of energy to change each kg of boiling water into steam.

How many joules of energy do you have available in the battery?

How does that help you answer the question that was asked.
 

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