Temperature of ice-water mixture

In summary, The ice cube in the beaker initially has a temperature of 0 degrees Celsius and a mass of 50g. When heated on a hot plate, half of the ice melts, meaning that the energy gained from the hot plate has gone to melting half of the ice's mass. This leaves the temperature of the ice-water mixture still at 0 degrees Celsius. When added to 2.5 kg of water at 25 degrees Celsius, the final temperature of the equilibrium mixture is 1.5 degrees Celsius.
  • #1
jiceo1
7
0

Homework Statement


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

Homework Equations


Q=mL
cmdT=Q


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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  • #2
Hi jiceo1,

What is true about ice-water mixtures that can help you know their temperature?
 
  • #3
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?
 
  • #4
I think that sounds right.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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?
 
  • #8
Not quite, I don't think; what is the initial and final temperature of the water as it warms?
 
  • #9
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
 
  • #10
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?
 
  • #11
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 statements about energy gained and energy lost sound good. But I'm not sure about the 1.5 degrees.
 
  • #12
[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
 
  • #13
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?
 

1. What is the temperature of an ice-water mixture?

The temperature of an ice-water mixture is typically at or near 0°C (32°F). This is the temperature at which water freezes and ice melts, creating a stable equilibrium.

2. Why does the temperature of an ice-water mixture remain constant?

The temperature of an ice-water mixture remains constant because the energy being absorbed by the ice as it melts is equal to the energy being released by the water as it freezes. This maintains a balance and keeps the temperature stable.

3. Can the temperature of an ice-water mixture change?

Technically, yes, the temperature of an ice-water mixture can change if there is a change in the surrounding environment or if there is a significant change in the ratio of ice to water in the mixture. However, under normal conditions, the temperature will remain constant.

4. What factors can affect the temperature of an ice-water mixture?

The temperature of an ice-water mixture can be affected by external factors such as the ambient temperature, humidity, and air flow. It can also be affected by the size and shape of the ice cubes and the amount of agitation in the mixture.

5. How can the temperature of an ice-water mixture be measured?

The temperature of an ice-water mixture can be measured using a thermometer. It is important to ensure that the thermometer is accurate and calibrated properly for an accurate reading.

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