Heat Exchange: Ice added to water

In summary: So the negative temperature is an indicator that the ice didn't all melt - so you need to adjust the mass of ice until you get a final temperature of 0o - which means that all the ice has melted.If you do that, the mass of ice you have used is the mass of ice remaining - which is the answer to part (b).
  • #1
XwyhyX
15
0

Homework Statement



A glass containing water is initially of temperature 20oC. The mass of the glass is 100 g and of specific heat 0.16 cal/g-co and water is of mass 200g. If 5 cubes of ice each of mass 15g will be added to the glass of water, solve for
(a) The temperature of the mixture.
(b) The mass of ice remaining if any.

Homework Equations



Qloss+Qgained=0
Q=mCΔt

The Attempt at a Solution



Ok so used the equation for heat exchange and I arrived at the equation,

mCΔt of water + mCΔt of glass + mCΔt of ice = 0

My concern is the Specific heat that I will use for C of ice, because my teacher said to only use 0.5 when below 0 degrees but it doesn't really say its temperature in the problem. Will I use 1 or 0.5?

And for part b, I need some tips on how to start solving for the possible remaining ice
 
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  • #2
well if you want to find the temperature of the mixture, you'll have to do the following:

[tex]0=\mu_{cup}(\theta_{f}-\theta_{i})+m_{1}C_{w}(\theta_{f}-\theta i)+m_{2}C_{ice}(\theta_{f}-\theta_{i})[/tex]

after you write that down you'll have to foil everything out then factor out what needs to be factored out with θf and just do simple algebra to find the final temperature.
 
  • #3
XwyhyX said:

Homework Statement



A glass containing water is initially of temperature 20oC. The mass of the glass is 100 g and of specific heat 0.16 cal/g-co and water is of mass 200g. If 5 cubes of ice each of mass 15g will be added to the glass of water, solve for
(a) The temperature of the mixture.
(b) The mass of ice remaining if any.


Homework Equations



Qloss+Qgained=0
Q=mCΔt

The Attempt at a Solution



Ok so used the equation for heat exchange and I arrived at the equation,

mCΔt of water + mCΔt of glass + mCΔt of ice = 0

My concern is the Specific heat that I will use for C of ice, because my teacher said to only use 0.5 when below 0 degrees but it doesn't really say its temperature in the problem. Will I use 1 or 0.5?

And for part b, I need some tips on how to start solving for the possible remaining ice

The ice is probably starting at 0o, so you don't have to worry about it warming up. You do have to account for the ice melting though - you need to use the latent heat of fusion.

Part (b) suggest not all the ice will be melted, so I would be working out how much heat energy has to be absorbed to cool the glass and water to 0o, and see how much ice that heat would melt.
Suppose that will melt 55 grams of ice, then the final mix will be at 0o, with 20g [of the original 75g,] still remaining.
 
  • #4
mtayab1994 said:
well if you want to find the temperature of the mixture, you'll have to do the following:

[tex]0=\mu_{cup}(\theta_{f}-\theta_{i})+m_{1}C_{w}(\theta_{f}-\theta i)+m_{2}C_{ice}(\theta_{f}-\theta_{i})[/tex]

after you write that down you'll have to foil everything out then factor out what needs to be factored out with θf and just do simple algebra to find the final temperature.

SO what about melting the ice!
 
  • #5
@PeterO

So you mean that instead of (mCΔt) of ice for the Qloss, i'll just have to use (mLatent) of ice instead and that only?
 
  • #6
XwyhyX said:
@PeterO

So you mean that instead of (mCΔt) of ice for the Qloss, i'll just have to use (mLatent) of ice instead and that only?

In most of these problems, yes.

Sometimes we are told that the ice cubes begin at a negative temperture - like -10o - and you then also need a bit of (mCΔt) of ice to allow for the ice warming up to 0o. If you are not given the beginning temperature of the ice, it is reasonable to assume it begins at 0o.
 
  • #7
I tried that, making the equation

mcΔt of water + mcΔt of glass +mLf

I used this for part a. but i get a negative value,-7.78 degrees is that possible, since i assumed that the ice started from 0 degrees it can't go lower than that can they?
 
  • #8
bro,

some part of the ice melts, and the other doesnt. madali lng yn homework :)
 
  • #9
blackandyello said:
bro,

some part of the ice melts, and the other doesnt. madali lng yn homework :)

Ano sagot? haha!
 
  • #10
XwyhyX said:
I tried that, making the equation

mcΔt of water + mcΔt of glass +mLf

I used this for part a. but i get a negative value,-7.78 degrees is that possible, since i assumed that the ice started from 0 degrees it can't go lower than that can they?

That negative temperature is an indicator that not all the ice melted. If you considered the ice had a smaller mass, the answer would have been zero degrees.
That leads to part (b)

re-read post #3; my first response.
 
  • #11
PeterO said:
That negative temperature is an indicator that not all the ice melted. If you considered the ice had a smaller mass, the answer would have been zero degrees.
That leads to part (b)

re-read post #3; my first response.

Thanks. I get it now. :D Last concern, is the negative temperature acceptable as the final temperature of the mixture? Just so I can clear things out. Haha

Thanks again! :D
 
  • #12
XwyhyX said:
Thanks. I get it now. :D Last concern, is the negative temperature acceptable as the final temperature of the mixture? Just so I can clear things out. Haha

Thanks again! :D

The final temperature will never be higher than the beginning highest temperature, nor lower than the beginning lowest temperature.

If we assume the ice begins at 0o that means the final temperature will be at, or above 0o.The only way to get a negative final temperature would be to begin with ice at a very negative temperature - and then the final mix would be solid [ice].
 

1. What happens to the temperature of water when ice is added?

When ice is added to water, the temperature of the water decreases. This is due to the transfer of heat from the water to the ice, causing the molecules in the water to slow down and lower the overall temperature.

2. Why does the temperature of the water decrease when ice is added?

The temperature of the water decreases when ice is added because of a process called heat exchange. Heat always flows from a warmer substance to a cooler substance, so the ice absorbs heat from the water, causing the temperature of the water to drop.

3. How long does it take for the ice to melt in the water?

The time it takes for the ice to melt in the water depends on several factors, such as the temperature and volume of the water, as well as the size and amount of ice. Generally, it takes a few minutes for small ice cubes to melt in a glass of water at room temperature.

4. Does the amount of ice added to the water affect the temperature change?

Yes, the amount of ice added to the water does affect the temperature change. The more ice added, the greater the decrease in temperature of the water. This is because there is a larger surface area for heat exchange to occur between the water and the ice.

5. Is it possible for the temperature of the water to rise after adding ice?

No, it is not possible for the temperature of the water to rise after adding ice. As mentioned before, heat always flows from a warmer substance to a cooler substance. Therefore, the ice will continue to absorb heat from the water until it reaches the same temperature, resulting in a decrease in the temperature of the water.

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