Calorimetrey Problem - Final Temperature of water

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SUMMARY

The calorimetry problem involves mixing 0.50 kg of water at 80°C with 0.050 kg of ice at -5.0°C in an insulated container. The final temperature of the system, after all the ice has melted, is 65°C. The key equations used include Q=mcΔT for heat transfer and Q=mL for phase changes, with specific heat values of 4186 J/kgK for water and 2100 J/kgK for ice, and a latent heat of fusion of 3.34 x 10^5 J/kg. The initial attempt at the solution yielded an incorrect final temperature of 68°C due to miscalculations in the heat transfer signs.

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  • Understanding of calorimetry principles
  • Familiarity with the specific heat capacity of water and ice
  • Knowledge of latent heat of fusion
  • Ability to solve equations involving heat transfer
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  • Study the calculations for specific heat and latent heat
  • Practice solving similar calorimetry problems
  • Learn about the conservation of energy in thermodynamic systems
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Calorimetrey Problem -- Final Temperature of water

Homework Statement



In an insulated container, 0.50kg of water at 80C is mixed with 0.050kg of ice at -5.0C. After a while, all the ice melts, leaving only the water. Find the final temperature T_f of the water. The freezing point of water is 0 C.

Homework Equations


Q=mcΔT
Q=mL (heat transfer in a phase change)
specific heat of water c = 4186 J/kgK
specific heat of ice c = 2100 J/kgK
latent heat of fusion for water = 3.34 x 10^5 J/kg

The Attempt at a Solution



Here's my attempt at a solution which came to 68 degrees C (the correct answer is 65C). I've checked my math, so I don't think it was a simple computational error. I think I'm making a mistake in terms of my signs perhaps when it comes to heat entering and exiting.

Anyway, here's what I did.

Q(water) + Q(ice) + m(ice)L = 0

(.5)(4186)(T-80) + (.05)(2100)(T+5) + (.05)(3.34x10^5)=0

I understand that in the case of Q(water), heat is exiting, and in the case of Q(ice) heat is entering. And since the ice melts, in the case of mL, heat is also entering. I guess I'm just not getting something crucial, here.

Any help would be greatly appreciated! Thank you.
 
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The melted ice is water. Determine the heat needed to warm up the ice to the melting point, then calculate with the specific heat of water.

ehild
 

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