Question #24009

1 Answer
Ed Hitchcock
Oct 27, 2017

128820 J, due to latent heat of vaporization. Note that the answer may be slightly different if your textbook uses a different value for #L_v#.

Explanation:

When liquid water changes temperature, the amount of energy needed to change that temperature is the Specific Heat Capacity #C#. For water this is #4.184J/(gC^o)#. But when water boils, it takes far more energy to transform from a liquid to a gas than just to change temperature as a liquid. This is called the latent heat of vaprozation.

The latent heat of vaporization for water is #2260J/g#. Which means it requires #2260J# of energy to vaporize each gram of water.

So if we have 57g of water already at 100C, then the heat needed is given by:

#Q = mL_v#

#Q = 57g * 2260J/g#

#Q = 128820 J#

I have seen different values of #L_v# published in different sources. I have seen 2256, 2258, 2264, and even 2230. The answer in your course will thus depend on the value that is used.

If the water started below 100C, we would need to calculate that energy as well, using the specific heat capacity (#4.184J/(gC^o)#) mentioned above.

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