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

  • Thread starter Thread starter DrMcDreamy
  • Start date Start date
  • Tags Tags
    Heat
AI Thread Summary
To convert 7 kg of ice at 0°C to steam at 100°C, the total heat required is calculated by considering three stages: heating the ice to water, melting the ice, and vaporizing the water. The specific heat of water is used to calculate the heat needed to raise the temperature, while the heat of fusion and vaporization are applied for phase changes. The calculations yield 700,000 cal for heating, 556,500 cal for melting, and 3,780,000 cal for vaporizing, resulting in a total of 5,036,500 cal, or 5036.5 kcal. This systematic approach confirms the correct application of thermodynamic principles for phase transitions. The final answer is 5036.5 kcal.
DrMcDreamy
Messages
68
Reaction score
0

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 ?
 
Physics news on Phys.org
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!
 
Back
Top