Calculating Final Temperature of Isolated System with Ice and Water

AI Thread Summary
The discussion revolves around calculating the final temperature of an isolated system containing 100g of ice at -15°C and 200g of water at 25°C. Participants emphasize the importance of considering both the specific heat capacities and the latent heat of fusion for the ice. The calculations reveal that the ice requires 3330J to reach 0°C, while the water can only provide 17620J before reaching 0°C, leading to a thermal equilibrium at 0°C with some ice remaining. The final consensus indicates that the system stabilizes at 0°C, confirming the calculations align with expected outcomes. Understanding the energy transfer and phase changes is crucial for solving such thermal equilibrium problems.
marcov8
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i have an easy problem that i can't solve

if you throuth 100g of ice at a tempeture of -15C in a glass with 200g of water at a tempeture of 25C.

what it the final tempeture of the system? the system is isolated, no heap can get in neither get out.

i try to solve this problem using specific heat capacity of water and ice, but i get weird result.
for water i used heat capacity = 4190 J/kg*K
heat capaciy of ice = 2220 J/Kg*K

can anyone give an idea on how to setup the equation?
thanks
 
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marcov8 said:
i have an easy problem that i can't solve

if you throuth 100g of ice at a tempeture of -15C in a glass with 200g of water at a tempeture of 25C.

what it the final tempeture of the system? the system is isolated, no heap can get in neither get out.

i try to solve this problem using specific heat capacity of water and ice, but i get weird result.
for water i used heat capacity = 4190 J/kg*K
heat capaciy of ice = 2220 J/Kg*K

can anyone give an idea on how to setup the equation?
thanks

Have you considered the latent heat involved when the ice melts?
 
Yes i consider that. but how do i know all the ice is melted?

if all the ice is melted

ICE:

Q_1 = cm(T_i - T_f)
Q_2 = Lm

for the ice the total heat Q_1 + Q_2

WATER:

Q = cm(T_i - T_f)

the final equation looks like

Q_1 + Q_2 = Q

then using the correct values of c,m and L i just solve for T_f

is this correct??
 
That looks about right but I would calculate the temperature difference as Tf -Ti. If you do it the other way round, a negative sign somewhere might lead you to the wrong answer.
 
oh yes that was a typo i mean to write T_f - T_i
 
Thanks all for your help, but i can't get the answer to my problem.
i'm using this values

ICE
<br /> $<br /> T_i = -15C = 258.15K \\<br /> m_i = 100g = 0.1Kg \\<br /> c_i = 2220 \frac{J}{Kg*K} \\<br /> L_i = 333 \frac{KJ}{Kg}<br /> $<br />

Water
<br /> $<br /> T_i = 25C = 298.15K \\<br /> m_w = 200g = 0.2Kg \\<br /> c_w = 4190 \frac{J}{Kg*K} <br /> $<br />

ok now, the equation for the ice:

<br /> $<br /> Q_1 = c_im_i(T_f-T_i) = c_i_m_iT_f-c_im_iT_i \\<br /> Q_2 = L_im_i<br /> $<br />

the equation for the water

<br /> $<br /> Q = c_wm_w(T_f-T_i) = c_wm_wT_f-c_wm_wT_i <br /> $<br />

no heat can get in neither get out so to get the final tempeture i solve for T_f the next equation

<br /> $<br /> Q_1+Q_2 = Q \\<br /> \\<br /> c_i_m_iT_f-c_im_iT_i + L_im_i = c_wm_wT_f-c_wm_wT_i \\<br /> \\<br /> T_f = \frac{c_im_iT_i-L_im_i-c_wm_wT_i}{c_im_i-c_wm_w}\\<br /> $<br />
i get $T_f = 366.62K$ wrong\\


Could you tell me where is my mistake??
 
Ok, there are 3 parts to the problem. The first part is from ice at -15 to ice at 0. Then there is ice at 0 to water at 0. Then, it's water at 0 with water at some other temp to be determined. I'm currently trying to work out the equations, will post when done.
 
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This seems to be a peculiar problem. My first guess is that you will end up with some ice and some water both at 0C. Is this an actual question from a book?

If you work it out step by step (I would write down equations but I've never used that LaTeX thing.):

First the ice takes heat from the water. It will require the ice 3330J to go from -15 to 0C.

When that heat is taken from the water, the new water temperature is 21.03C.

Now, if the ice is to completely melt, it will need 33300J of energy but the water at 21.03C can only supply 17620J of energy before it itself gets to 0C.

Therefore you end up with some ice/water equilibrium at 0C.

PS: My calculations are based on the specific heat capacities and latent heat that you have used.
 
  • #10
I work out that you end up with 47.09g of ice and 252.91g of water in equilibrium. Can anybody confirm this? It's been so long since I did heat calculations so I might have done silly mistakes somewhere.
 
  • #11
yes the answer on the back of the book is 0C.

Now, if the ice is to completely melt, it will need 33300J of energy but the water at 21.03C can only supply 17620J of energy before it itself gets to 0C.

thanks for your post now i know how calculated if the ice is melted or not.
i don't know why i didn't think of it before.

now i'll try to write down my equations
 
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