Calculating Heat Capacity, deltaU and deltaH of Cu Metal

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SUMMARY

The discussion focuses on calculating the heat capacity, change in internal energy (deltaU), and change in enthalpy (deltaH) for 100 grams of copper (Cu) heated from 0°C to 100°C at constant pressure (1 atm). The molar heat capacity of Cu is 24.4 J/molK, leading to a calculated Q and deltaH of approximately 3840 J. The work done (W) during this process is calculated as -0.0057 J, with the note that deltaH is approximately equal to deltaU for solids and liquids due to minimal volume changes during heating.

PREREQUISITES
  • Understanding of thermodynamic concepts such as heat capacity and enthalpy
  • Familiarity with the ideal gas law and pressure-volume work calculations
  • Basic knowledge of molar mass and density calculations
  • Proficiency in using the formula Q = nC(Tf - Ti)
NEXT STEPS
  • Study the relationship between deltaH and deltaU for different states of matter
  • Learn about the implications of temperature independence of heat capacities
  • Explore advanced thermodynamic calculations involving phase changes
  • Investigate the effects of pressure on heat capacity in different materials
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Chemistry students, thermodynamics enthusiasts, and professionals involved in material science or engineering who are looking to deepen their understanding of heat transfer and energy calculations in solids.

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Homework Statement


At 25C Cu metal (at. wt. 63.54 g/mol) has a molar heat capacity of 24.4 J/molK. The density of Cu is 8.949g/cm^3 at 0C and 8.904 g/cm^3 at 100C. Assuming that Cp is temperature independent, calculate Q, W, deltaU and deltaH when 100grams of Cu metal are heated from 0C to 100C under constant pressure 1atm.

Ans. Q=deltaH is about equal to deltaU = 3840J, W=-.0057J

Note: deltaH is about equal to deltaU for heating of solids and liquids at constant pressure because these materials exhibit very low volume changes on heating.

The Attempt at a Solution



density = m/v so v=m/density so
V1=100g/8.949g/cm^3 = 11.17cm^3 and V2 = 100g/8.904 g/cm^3 = 11.23cm^3
W=-P(V2-V1) = -1atm((11.23X10^-2(m^3))-(11.17X10^-2(m^3)))=-.0006J about correct?

Q=nCn(Tf-Ti)
n=((63.54g/mol)/(100g))=.6354mol
Q = (.6354mol)(24.4J/molK)(373K) = 5782.9J not so correct?...
 
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If you have 100g and molar mass is 63.5g, having 0.63 moles is off. It is just like stating "I have $100, this book is $50, so I can buy only a half"...
 

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