Heat Exchange: Ice added to water

[XwyhyX]

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

Q_loss+Q_gained=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

 

[mtayab1994]

well if you want to find the temperature of the mixture, you'll have to do the following:

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

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.

 

[PeterO]

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

Q_loss+Q_gained=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 0^o, 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 0^o, and see how much ice that heat would melt.
Suppose that will melt 55 grams of ice, then the final mix will be at 0^o, with 20g [of the original 75g,] still remaining.

 

[PeterO]

well if you want to find the temperature of the mixture, you'll have to do the following:

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

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!

 

[XwyhyX]

@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?

 

[PeterO]

@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 -10^o - and you then also need a bit of (mCΔt) of ice to allow for the ice warming up to 0^o. If you are not given the beginning temperature of the ice, it is reasonable to assume it begins at 0^o.

 

[XwyhyX]

I tried that, making the equation

mcΔt of water + mcΔt of glass +mL_f

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?

 

[blackandyello]

bro,

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

 

[XwyhyX]

bro,

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

Ano sagot? haha!

 

[PeterO]

I tried that, making the equation

mcΔt of water + mcΔt of glass +mL_f

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.

 

[XwyhyX]

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

 

[PeterO]

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 0^o that means the final temperature will be at, or above 0^o.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].