Equilibrium Temperature of Cup and Water When Adding Ice - Homework Solution

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SUMMARY

The equilibrium temperature of a system consisting of a 35 g ice cube at 0.0°C, 110 g of water, and a 62 g iron cup initially at 36°C can be calculated using the heat transfer equations. The heat lost by the water and cup is represented by Q(w) = m(water)c(w)ΔT + m(cup)c(c)ΔT, while the heat required to melt the ice is given by Q(ice) = m(ice)L(f). The final equilibrium temperature is determined by the equation ΔT = Q/((m(w) + m(ice))c(w) + (m(cup))c(cup)). A miscalculation in unit conversion was identified as the source of error in the initial attempt to solve the problem.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Familiarity with specific heat capacity (c) values
  • Knowledge of latent heat of fusion (L(f))
  • Ability to perform unit conversions accurately
NEXT STEPS
  • Review the concept of heat transfer in thermodynamic systems
  • Study specific heat capacity calculations for different materials
  • Learn about latent heat and its applications in phase changes
  • Practice unit conversion techniques in physical chemistry problems
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Homework Statement


A 35 g ice cube at 0.0°C is added to 110 g of water in a 62 g iron cup. The cup and the water have an initial temperature of 36°C. Find the equilibrium temperature of the cup and its contents.


Homework Equations


heat lost by water+cup Q(w)=m(water)c(w)deltaT+m(cup)c(c)deltaT
amt. of heat needed to melt ice Q(ice)=m(ice)L(f)
amt. of heat left Q=Q(w)-Q(ice)
deltaT=Q/((m(w)+m(ice))c(w)+(m(cup))c(cup))

The Attempt at a Solution


I tried plugging in my numbers to the equations above but I didn't get the right answer. I'm fairly certain that my equations are correct but the only one that I am uncertain about is the last one.
 
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Never mind...I found out that it was simply a miscalculation error in converting units. :)
 

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