Specific heat capacity - heat energy transfer

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SUMMARY

The discussion centers on calculating the mass of water that vaporizes when 3.39 kg of mercury at 243 °C is introduced to 0.476 kg of water at 90.0 °C. The heat transfer equations used include q=mcΔT and q=mL, where the heat lost by mercury equals the heat gained by water. The user attempts to solve for the mass of vaporized water but arrives at an incorrect result of 2.1 x 10^-5 kg, questioning the derivation of the temperature difference of 100-10 °C.

PREREQUISITES
  • Understanding of specific heat capacity and the equation q=mcΔT
  • Knowledge of phase change and latent heat, specifically q=mL
  • Familiarity with temperature scales and conversions
  • Basic algebra skills for solving equations
NEXT STEPS
  • Review the concept of heat transfer in phase changes, focusing on latent heat calculations
  • Study the derivation and application of the specific heat capacity formula q=mcΔT
  • Learn about the properties of water and mercury, including their specific heat capacities
  • Practice similar problems involving heat transfer between different substances
USEFUL FOR

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

slaw155
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Homework Statement



Find the mass of water that vaporizes when 3.39 kg of mercury at 243 °C is added to 0.476 kg of water at 90.0 °

Homework Equations


q=mcT, q=mL


The Attempt at a Solution


When vaporising T=100degC. Heat lost by mercury = heat gained by water. 3.39kg x 139J/kg/C x (243-100) = 0.476kg x 4186J/kg/C x (100-10) + 22.6x10^5 J/kg x m
m= 2.1 x 10^-5kg is what I get which is the wrong answer. Where have I gone wrong?
 
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Where did the 100-10 come from?

Chet
 

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