## Temperature of ice-water mixture

1. The problem statement, all variables and given/known data
A beaker containing 50g of ice and 250g of liquid water is initially at 0celsius. It is then heated on a hot plate until half of the ice is melted. What is the temperature of the ice-water mixture at that point?
A. 2.0 C
B. 0.5 C
C. 1.0 C
D. 1.5 C
E. 0.0 C

2. Relevant equations
Q=mL
cmdT=Q

3. The attempt at a solution
I'm still not sure about how to set up the equations at this starting point.
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 Recognitions: Homework Help Hi jiceo1, What is true about ice-water mixtures that can help you know their temperature?
 I assume that since half of the ice is still there in the water, then the energy gained from the hot plate has all gone to melting half of the ice's mass. So at this point, the temperature would still be 0 degrees, am I wrong?

Recognitions:
Homework Help

## Temperature of ice-water mixture

I think that sounds right.
 I have another related question, so I'm going to post it here instead of creating a new topic: An ice cube has a mass of 720 g and a temperature of -10° C. Assume that the specific heat of ice is c_ice = 2220 J/kg•°C, the latent heat of fusion of ice is L_F = 333 KJ/kg and the specific heat of liquid water is c_water = 4190 J/kg•°C. a. The ice cube melts and comes to a final temperature of 15° C. How much heat was added to the ice cube? Total energy= mL_F + cmdT (0.720kg)(333kJ/kg) + (4.190kJ/kgC)(0.720kg)(15-(-10)) 239.76 +75.42 = 315.18 kJ energy, is this correct? b. The ice cube is added to 2.5 kg of water at 25° C. What is the final temperature of the equilibrium mixture? c. If 720 g of ice water is added instead to the 2.5 kg of water at 25° C, what is the final equilibrium temperature of the mixture?
 Recognitions: Homework Help The specific heat of ice and water are not the same, so ice and water will need separate terms in the equation. For part a there are three processes that occur: first the ice warms up to the freezing point, then it melts, then the water warms up to the final termperature.
 oh, ok: so for part A it would be: (2220 J//kgC)(0.720kg)(0-(-10)) + (0.720)(333000 J/kg) + (4190 J/kgC)(0.720kg)(15-(-10)) = 331000 J correct?
 Recognitions: Homework Help Not quite, I don't think; what is the initial and final temperature of the water as it warms?
 ooh, I see it... as the water warms up, it warms from 0 degrees to 15 degrees. -_- my bad... so my final answer is: 300,996 J
 for part B, since energy gained = energy lost. Energy gained = the ice warming up to 0, then melting, then the water at 0 degrees warming up to a final temperature. Energy lost = water cooling down to a final unknown temperature. I got 1.5 degrees for the final temperature. Correct?
 Recognitions: Homework Help If you don't mind, please post the numbers that you use in your calculation (the way you did in post #7). It makes it much easier to see what you did. Your statments about energy gained and energy lost sound good. But I'm not sure about the 1.5 degrees.
 [2220 J/kgC * 0.720kg * 0-(-10)] + [0.720kg * 333000 J/kg * 4190J/kgC * 0.720kg * (T-0)] = 4190 J/kgC * 2.5kg * (25-T) T=1.5 degrees
 Recognitions: Homework Help You have a small typo right after the 333000; it should be a plus sign, but you probably entered into your calculator correctly. So when I check the number by calculating the left side with T=1.5, I get: [2220 J/kgC * 0.720kg * 0-(-10)] + [0.720kg * 333000 J/kg + 4190J/kgC * 0.720kg * (T-0)] =15984+239760 + 4525.2 = 260269 and the right side is 4190 J/kgC * 2.5kg * (25-T) = 4190 * 2.5 (25-1.5) = 246163 Since the right and left sides don't equal, that's why I thought 1.5 degrees was not correct, and I think you made an algebraic error when solving for T. Combing the multiplications in your expression on the left side and multiplying out the terms on the right side gives: 15984+239760 +3016.8 T =261875-10475 T which you can then solve for T. Is that what you are getting?
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