How much heat is required to convert ice into steam?

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SUMMARY

The discussion focuses on calculating the heat required to convert an ice cube of volume 12.1 cm³ at -18.1°C into steam. The participant initially used a density of 9.17 kg/cm³ to find the mass, resulting in an incorrect total heat calculation of 5.09e14 Joules. Key errors identified include the incorrect density of ice and unit inconsistencies in the latent heat values for fusion and vaporization. Correcting these factors is essential for accurate heat calculations.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat transfer concepts.
  • Familiarity with the specific heat capacities of ice and water.
  • Knowledge of latent heat of fusion and vaporization for water.
  • Ability to convert between units, particularly from grams to kilograms.
NEXT STEPS
  • Review the specific heat capacity of ice and water, focusing on values of 2090 J/kg°C and 4180 J/kg°C.
  • Learn about the latent heat of fusion (334,000 J/kg) and vaporization (2,200,000 J/kg) for water.
  • Practice unit conversion techniques, especially between grams and kilograms in thermodynamic calculations.
  • Explore density variations of substances, particularly the density of ice compared to water.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, educators teaching heat transfer concepts, and anyone involved in physics or engineering calculations related to phase changes of water.

theskyisgreen
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Homework Statement


An ice cube of volume 12.1 cm3 is initially at a temperature of -18.1°C. How much heat is required to convert this ice cube into steam?
(the final answer must be in Joules)


Homework Equations


Q = m*c*deltaT
Q = m*Lf (latent heat of fusion)
Q = m*Lv (latent heat of vaporization)
density = m/v


The Attempt at a Solution


Since I was only given the volume of the ice cube, I decided to use density = mass/volume to find the mass of the ice cube. Since the final answer must be in joules, I used the density of 9.17 kg/cm^3 as the density, and found the mass to be 1109570.

I then plugged in that mass into
Q = (1109570) (2090 J/kg°C the c of ice) (18.1) to convert ice to water
Q = (1109570) (4180 J/kg°C the c of water) (100) to convert water to steam
Q = (1109570) (334000 J - the Lf of water)
Q = (1109570) (2200000 J - Lv of water)

I added all the Qs together and found the final answer to be 5.09e14, which is incorrect.
Finding the correct units was a bit tricky so I may have done something wrong there.
If anyone has any suggestions, please let me know.
 
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Your approach is correct.

Double check the units and value for the density of ice. Recall the density of water is about 1 gr per cubic cm, so the density of ice should be slightly less than that.

Also, you've mixed units in the Lf and Lv for water. It looks like you might have switched from "cal / g degree' to 'J / g degree'. Don't forget to also change the g to kg.

Make sure everything is in the same units - m, kg, and J are all MKS units, so stick with those.
 

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