Heat energy transferred in converting ice into steam

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SUMMARY

The calculation of heat energy required to convert 0.10 kg of ice at -10 degrees Celsius into steam at 100 degrees Celsius involves multiple stages: heating ice to 0°C, melting ice to water, heating water to 100°C, and vaporizing water to steam. The correct total energy calculated is 301.6 kJ, which includes 2,100 J for heating ice, 32,500 J for melting, 42,000 J for heating water, and 225,000 J for vaporization. The error in the initial calculation stemmed from not accounting for the latent heat of vaporization correctly, which is 2,300,000 J/kg. This comprehensive approach confirms that the majority of energy is required for the phase change from water to steam.

PREREQUISITES
  • Understanding of specific heat capacity and phase changes
  • Familiarity with the concepts of latent heat of fusion and vaporization
  • Basic knowledge of thermodynamics and energy calculations
  • Ability to perform unit conversions between Joules and kilojoules
NEXT STEPS
  • Study the concept of specific heat capacity in detail
  • Learn about latent heat of fusion and vaporization for different substances
  • Explore thermodynamic processes involving phase changes
  • Practice energy calculations involving multiple phase transitions
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Students in physics or chemistry, educators teaching thermodynamics, and anyone interested in understanding energy transfer during phase changes of water.

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1. Homework Statement

Calculate the heat energy required to convert 0.10 kg of ice at -10 degrees celsius into steam at 100 degrees celsius.




2. Homework Equations

E = m c dt (1)

where

E = heat energy (kJ)

m = unit mass (kg)

c = specific heat capacity (kJ/kg oC)

dt = temperature change (Kelvin or degrees Celsius)





3. The Attempt at a Solution

Given in the question:

m = 0.1 kg
dt = 100-(-10) = 110 degrees celsius
c = 2100 J Kg^-1 K^-1 (specific heat capacity of ice)


So what I did was using the equation:

E = m c dt

E = (0.1)(2100)(110)

E = 23100 J

which would be 23.1 kJ



The problem is that the answer section of the book says 301.6 KJ which is way above my result.
What's the mistake I'm doing?

Help appreciated.

 
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Ice must be heated to 0C; ice must then be converted into liquid water at 0C;
Water at0C must be heated to 100C; water must be converted to vapor (steam) at 100C.
There are 4 stages, each requiring heat energy
 
Thanks for the hint.For Ice -> water:

m = 0.1kg c=2100 J kg^-1 K^-1 dt = 10 degrees (-10 -> 0)

therefore E= mc dt = 0.1(2100)(10) = 2100 J

Energy required to melt the ice:

E= specific latent heat of fusion of ice * mass
= 325000(0.1)
= 32500 J

and then for water to steam:

m = 0.1 kg c= 4200 J kg^-1 K^-1 (for water) dt= 100 degrees (0->100)

=> E= 0.1 (4200) (100) = 42000 J

Energy required to vaporize 0.1 kg of water:

E = specific latent heat of vaporization of water * mass
= 225000(0.1)
= 22500 J

then i add them

2100 + 32500 + 42000 + 225000

and get 301600 J which is 301.6 kJ.Thank you.
 
That looks good.
Well done
Do you see that by far the greatest amount of energy is in converting liquid water to steam?
 
SORRY... misprint, you have the latent heat of vaporisation to convert water to steam out by factor of 10
Latent heat of vaporisation = 2300000 J/kg
I have only just noticed... fortunately you made a mistake in your multiplication so your answer is correct6.
should always double check calculations !
 

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