How much heat is needed for converting 7 kg of ice to steam?

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Homework Help Overview

The problem involves calculating the total heat required to convert 7 kg of ice at 0°C into steam at 100°C. This encompasses the phases of melting ice to water and then vaporizing water to steam, utilizing specific heat capacities and latent heats.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to apply different formulas for each phase change, including the heat of fusion and vaporization. There are questions about the correct formulas to use and the process of summing the heat values for each stage.

Discussion Status

Some participants have offered guidance on the formulas to use, while others have confirmed the approach of calculating heat for each phase separately. There is an ongoing exploration of the calculations involved, but no explicit consensus has been reached regarding the final answer.

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Participants are working within the constraints of a homework assignment, which may limit the depth of discussion and the information shared.

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Homework Statement



7 kg of ice at 0◦C is converted to steam at 100◦C. How much heat is needed? The heat of fusion for water is 79500 cal/kg, its heat of vaporization is 5.4 × 105 cal/kg, and its specific heat is 1000 cal/kg ·◦C. Answer in units of kcal.

Homework Equations



What is the formula that I am supposed to use?

Q=mc\DeltaT ?
 
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You *also* need to use

Q = mL,
where m is the mass and L is the heat of fusion/vaporisation.R.
 
you go from solid to liquid to gas, so you have to calculate the heat for each stage than add up the total of each stage. I think, also, for the vaporization you would use Q =ml where L = latent heat of vaporization.. and Q=ml again for latent heat of fusion/melting

I think that's how you do it...
 
Last edited:
You should end up with 3 separate values for Q which you will have to add up.
 
So:

Q=mc\DeltaT
Q=(7kg)(1000 cal/kg C)(100 C)
Q= 700,000 cal

heat of fusion:
Q=mL
Q=(7kg)(79500 cal/kg)
Q= 556,500 cal

heat of vaporization:
Q=mL
Q=(7kg)(5.4x105 cal/kg)
Q=3,780,000 cal

\sumQ= 700,000 + 556,500 + 3,780,000

\sumQ=5,036,500 cal = 5036.5kcal

Is that how?!
 
Indeed :)
 
Thank you!
 

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