Ice Cubes and Drink Equilibrium

1. The problem statement, all variables and given/known data

In an attempt to quickly make iced tea, three 50 g ice cubes with a temperature of -10 C are added to 250 g of tea with a temperature of 50 C. After a short time, the system reaches equilibrium. Ice has a specific heat capacity of 0.48 kcal/kg/K and the latent heat of fusion for ice/water is 80 kcal/kg. Assume that tea has the same thermal properties as water, and ignore any heat exchanged with the environment.

1. What equilibrium temperature does the tea reach?
2. Does all of the ice melt? If not, how much is left?

2. Relevant equations

Q=mL and Q=cmΔT

3. The attempt at a solution

I have reached a solution, but conceptually don’t understand it fully, and thus I’m not sure if it is right nor am I sure my answer is the "finished" answer. I have left the units in kcal/kg and C since I do not believe it makes a difference, as the differences in C are equivalent to those in K. As well, I figured if I kept everything in kcal it wouldnt make a difference between kcal or J.

To start, I found the decrease in temperature of the tea to first reach the ice’s melting point.
Since heat must be conserved, Qice=Qtea

(0.150kg)*(0.48kcal/kg/C)*(10°C) = (0.250kg)*(1kcal/kg/C)*(50°C-Final Temperature)

Final Temperature = 47.12°C

From there, I tried to find how much ice melts. At first I simply did Qice=Q tea where mL = cmΔT
but found that the tea ended up at -1.2°C. Thus, all the ice couldn’t have melted. Therefore, I tried to find how much DID melt.

m*(80kcal/kg/c) = (0.250g)*(1kcal/kg/C)*(47.12°C-0°C)
m= 0.147kg ice
thus 0.150kg – 0.147kg ice dictates that 0.003kg of ice was left over, or 3g of ice.

Thus I concluded that 1) the equilibrium temperature must be 0°C and 2) there is 3g of ice left floating around. I figured this was the end, but I’m not sure. Does the ice just stick around and no more heat is exchanged? I’m really confused conceptually for this specific instance.

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