Calculating Heat Required to Change 1.0 kg of Ice to Steam

AI Thread Summary
To calculate the heat required to convert 1.0 kg of ice at -30 degrees Celsius to steam at 120 degrees Celsius, the process must be divided into five stages: heating the ice, melting it, heating the water, vaporizing it, and heating the steam. The relevant constants include the latent heat of fusion (79.7 kcal/kg), latent heat of vaporization (539 kcal/kg), and specific heats for ice (0.5 kcal/kg/K), water (1.0 kcal/kg/K), and steam (0.480 kcal/kg). The initial attempt used an incorrect formula and values, leading to confusion about the pressure's role, which is not directly involved in the calculations. A structured approach using the specified equations for each phase of the transition is essential for an accurate solution. Understanding these steps will clarify the calculation process.
Mr. Goosemahn
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Homework Statement


How much heat in kcal is required to change 1.0 kg of ice, originally at -30 degrees C, into steam at 120 degrees C? Assume 1.0 atm of pressure. (Please use the following values for the constants needed in the problem: Latent heat of fusion=79.7 kcal/kg; Specific heat of ice=0.5 kcal/kg/K; Latent heat of vaporization=539 kcal/kg; Specific heat of water 1.0 kcal/kg/K; Specific heat of steam=0.480 kcal/kg)


Homework Equations


Qmelt=mLfusion
Qfreeze=-mLfusion
Qvaporization=mLvaporization
Qcondensation=-mLvaporization


The Attempt at a Solution


Quite honestly, I don't even know how to start. I don't see how the 1.0 atm of pressure is involved in the equations.

Could somebody just tell me how to start this off and I'll take it from there.

I've been reading from my book but it's not clear.

I had originally tried to use the equation Q=cmT, with the values plugged in as Q=4186*1*150, giving me 627.9 kilocalories, but this is incorrect.
 
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Divide the problen into five sections.
i) Q1 = m*ci*ΔT----(1)
ii) Q2 = m*Lfreeze -----(2)
similarly right the remaining equations and find the total Q..
 

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