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

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SUMMARY

To convert 7 kg of ice at 0°C to steam at 100°C, a total heat of 5,036.5 kcal is required. This calculation involves three stages: heating the ice to 0°C, melting the ice to water, and vaporizing the water to steam. The specific heat of water is 1,000 cal/kg·°C, the heat of fusion is 79,500 cal/kg, and the heat of vaporization is 540,000 cal/kg. The total heat is calculated by summing the heat required for each stage: 700,000 cal for heating, 556,500 cal for fusion, and 3,780,000 cal for vaporization.

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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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