Calculating Final Temperature of Isolated System with Ice and Water

In summary, the problem is to determine the final temperature of a system when 100g of ice at -15C is added to 200g of water at 25C in an isolated system. The specific heat capacities of water and ice are given, but the calculation yields a strange result. After considering the latent heat involved when the ice melts, the correct equation is set up to find the final temperature. However, due to the limited heat capacity of the water, the final temperature is found to be 0C, resulting in a mixture of ice and water in equilibrium.
  • #1
marcov8
5
0
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
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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??
 
  • #4
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.
 
  • #5
oh yes that was a typo i mean to write T_f - T_i
 
  • #7
Thanks all for your help, but i can't get the answer to my problem.
i'm using this values

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

Water
[tex]
$
T_i = 25C = 298.15K \\
m_w = 200g = 0.2Kg \\
c_w = 4190 \frac{J}{Kg*K}
$
[/tex]

ok now, the equation for the ice:

[tex]
$
Q_1 = c_im_i(T_f-T_i) = c_i_m_iT_f-c_im_iT_i \\
Q_2 = L_im_i
$
[/tex]

the equation for the water

[tex]
$
Q = c_wm_w(T_f-T_i) = c_wm_wT_f-c_wm_wT_i
$
[/tex]

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

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


Could you tell me where is my mistake??
 
  • #8
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.
 
Last edited:
  • #9
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
 

Related to Calculating Final Temperature of Isolated System with Ice and Water

What is thermodynamics?

Thermodynamics is the branch of science that deals with the relationship between heat, energy, and work. It explains the behavior of matter and its ability to transfer and transform energy.

What is a thermodynamic problem?

A thermodynamic problem is a question or challenge that requires the application of thermodynamic principles to solve. It may involve calculating energy transfers, changes in temperature or pressure, or determining the equilibrium state of a system.

What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern energy and its transformations in a physical system. They include the laws of conservation of energy, entropy, and the direction of heat flow.

What are some real-world applications of thermodynamics?

Thermodynamics has many practical applications, including power generation, refrigeration and air conditioning, chemical reactions, and engines. It also plays a crucial role in understanding weather patterns and climate change.

How does thermodynamics relate to other branches of science?

Thermodynamics is a fundamental aspect of many scientific disciplines, including physics, chemistry, engineering, and biology. It helps explain the behavior of matter at a molecular level, making it essential to understanding many natural phenomena and technological processes.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
4K
  • Introductory Physics Homework Help
Replies
11
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
965
  • Introductory Physics Homework Help
Replies
4
Views
709
Back
Top