Calculating Heat Required for Ice Cube Transformation

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SUMMARY

The calculation of heat required to transform a 46.6 g ice cube from -12.5°C to water at 53°C involves multiple steps and specific heat capacities. The correct specific heat capacity for water is 4180 J/kg°C, not 2010 J/kg°C, which is for steam. The total heat required is calculated using the formulas Q = mcΔT and Q = mL, resulting in a total of 21792.723 J when the correct values are applied. The error in the original calculation stemmed from using the wrong specific heat capacity for the final phase of water.

PREREQUISITES
  • Understanding of specific heat capacity and phase changes
  • Familiarity with the formulas Q = mcΔT and Q = mL
  • Knowledge of the specific heat capacities of ice and water
  • Basic principles of thermodynamics
NEXT STEPS
  • Study the specific heat capacities of different substances, focusing on water and ice
  • Learn about phase change calculations in thermodynamics
  • Explore the concept of latent heat and its applications
  • Practice problems involving heat transfer and temperature changes in various materials
USEFUL FOR

Students in chemistry or physics, educators teaching thermodynamics, and anyone interested in understanding heat transfer and phase changes in materials.

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How much heat is required to change a 46.6 g ice cube from ice at -12.5°C to water at 53°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C)

then i used Q=cmt and Q=mL

Q=(2090)(.0466)(12.5)=1217.425J
Q=(.0466)(33.5E4)=15611J
Q=(2010)(.0466)(53)=4964.298J

then when i add all the numbers up i get 21792.723J total but the answer is wrong. i don't know if I am using the formulas in the wrong order or what. I've done these problems in chemistry but its been a couple years since i took one of those courses. any help is appreciated.
 
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You use C_water, not C_steam for the last calculation (C_water is 4180)
 

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