Thermodynamics Homework Help: Finding Final Temperature in a Water-Ice Mixture

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To find the final temperature of a water-ice mixture in thermodynamics, the heat exchanges must be accurately accounted for. The initial setup includes 0.50 kg of water at 80°C and 0.050 kg of ice at -5°C. The calculations should consider three stages: heating the ice from -5°C to 0°C, melting the ice, and then heating the resulting water to the final temperature T_f. The specific heat capacities of ice and water differ, which affects the calculations for Q_ice. Correcting the approach by treating the ice in these three parts will yield the accurate final temperature.
Corneo
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I have a question about my Thermodynamics homework.

In an insulated container, 0.50 kg of water at 80^\circ C is mixed with 0.050 kg of ice at -5.0^\circ C. 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^\circ C

I need to find all the Q's and set their sum to equal 0.

I have the Q for the water, Q for the ice, Q from fusion. T_f is the same for all the substances. But my answer is still off.

Q_{H_2 0}= 0.5 kg c_{H_2 0} ( T_f-80^\circ C )
Q_{ice}= 0.05 kg c_{ice} (T_f+5^\circ C)
Q_{f}= 0.05 kg L_f

Is there another Q I haven't accounted for?
 
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what are the specific heat capacity of ice and water in your question? Are they being stated to be the same or different? If different, then you have to recheck your Q_{ice}.
 
They are different. What is wrong with my Q_{ice}?
 
Corneo said:
Q_{ice}= 0.05 kg c_{ice} (T_f+5^\circ C)
Your error is treating the ice as if it remained ice. You need to treat it in three parts:
(1) heating of 0.05 kg of ice from -5 to 0
(2) melting of ice (you have that piece already)
(3) heating of 0.05 kg of water from 0 to T_f
 
Ah... thank you very much.
 
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