[Thermodynamics] Impossible Heat Transfer Problem

[Gr33nMachine]

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??

 

[256bits]

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.

 

[Chestermiller]

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.