[Thermodynamics] Impossible Heat Transfer Problem

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SUMMARY

The forum discussion addresses a thermodynamics problem involving heat transfer between water and ice in an insulated Thermos. The initial calculations for the final temperature (77.63K) are incorrect due to the improper use of heat capacities. Specifically, the heat capacity of liquid water must be used in both terms of the denominator after the ice has melted. The latent heat of fusion (L) for ice is correctly identified as 334 J/g, which is essential for calculating the energy required to melt the ice.

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Gr33nMachine
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An insulated Thermos contains 122 g of water at 91°C. You put in a 13.7 g ice cube at 0°C to form a system of ice + original water.

Use mc(ΔT) |water + mc(ΔT) |ice + Lm |ice = 0

Final T = [(m*c*T) |water + (m*c*T) |ice - (L*m) |ice] / [m*c |water + m*c |ice]

Final T = [(.122*4.18*364) + (.0137*2.108*273) - (334*.0137)] / [.122*4.18 + .0137*2.108]

[STRIKE]Final T = 77.63K[/STRIKE]
IS WRONG

What am I doing wrong??
 
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assuming you went from
Use mc(ΔT) |water + mc(ΔT) |ice + Lm |ice = 0
to
Final T = [(m*c*T) |water + (m*c*T) |ice - (L*m) |ice] / [m*c |water + m*c |ice]

What is L and why is it 334.
 
Gr33nMachine said:
An insulated Thermos contains 122 g of water at 91°C. You put in a 13.7 g ice cube at 0°C to form a system of ice + original water.

Use mc(ΔT) |water + mc(ΔT) |ice + Lm |ice = 0

Final T = [(m*c*T) |water + (m*c*T) |ice - (L*m) |ice] / [m*c |water + m*c |ice]

Final T = [(.122*4.18*364) + (.0137*2.108*273) - (334*.0137)] / [.122*4.18 + .0137*2.108]

[STRIKE]Final T = 77.63K[/STRIKE]
IS WRONG

What am I doing wrong??

In the denominator, you should be using the heat capacity of liquid water in both terms, since the ice is melted.
 

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