Entropy-any direction would be appreciated

[hils0005]

A cup made of 100g of aluminum holds 200g of water at 50deg C, a 10g piece of ice is placed in the cup of water, what is the final temperature?

2. Homework Equations
Q=L(f)m
S=cm lnTf/Ti
specific heat of aluminum 0.90 J/gC

The Attempt at a Solution

I think the first step is to determine S(ice)
Q=Lm=79.5 cal/g * 10g=795cal
S(ice)=Q/T=795/273K=2.91
S(water)=Q/T=-795/323K=-2.46
2.91-2.46=.45cal/K

then as 0deg C water goes to 50deg C?
I believe I should be using an integral as the ice will be changing the temp of the water?
S(ice)=cmln(Tf/Ti)
1cal/gC * 10g ln(795/323)=9

S(water)=cmln(Tf/Ti)
1cal/gC*200 ln(795/323)=180

So now I'm confused-because I believe I would need to know the final temp of the water before calculating S with the Steel??
I also do not understand how to convert J/gC to cal

 

[Andrew Mason]

A cup made of 100g of aluminum holds 200g of water at 50deg C, a 10g piece of ice is placed in the cup of water, what is the final temperature?

2. Homework Equations
Q=L(f)m
S=cm lnTf/Ti
specific heat of aluminum 0.90 J/gC

This does not involve an entropy calculation. This is simply a heat flow calculation. You have to know the specific heat of Al, Water and Ice as well as the heat of fusion of water/ice. Then you have to set up equations for the change in temperature in terms of the heat flow from the water/cup to the ice and the final temp. Then solve the equations.

You can't convert Joules/g C to Calories. You can convert Joules/g C to Calories/g C by dividing by 4.187 Joules (= 1 Cal.)

AM

 

[Moetasim]

/gC.

I can provide some guidance on how to approach this problem. First, let's define entropy. Entropy is a measure of the disorder or randomness in a system. In simpler terms, it is a measure of how spread out or dispersed the energy is within a system. In this case, we can think of the cup of water as our system.

To solve this problem, we need to consider the changes in entropy that occur when the ice is added to the cup of water. The first step is to calculate the change in entropy of the ice. We can use the equation S=Q/T, where Q is the heat absorbed by the ice and T is the temperature at which the heat is absorbed. We know that the heat absorbed by the ice is equal to its latent heat of fusion, which is 79.5 cal/g. The temperature at which this heat is absorbed is the melting point of ice, 0°C or 273K. So, the change in entropy of the ice is 79.5 cal/g / 273K = 0.291 cal/K.

Next, we need to calculate the change in entropy of the water. This will involve a temperature change from 50°C to the final temperature, T. We can use the equation S=cm ln(Tf/Ti), where c is the specific heat of water (1 cal/g°C) and m is the mass of water (200g). So, the change in entropy of the water is 200g * 1 cal/g°C * ln(T/Ti). We can also rewrite this equation as S=Q/T, where Q is the heat absorbed by the water and T is the temperature at which the heat is absorbed. The heat absorbed by the water is equal to the heat released by the ice, which we calculated to be 79.5 cal/g. So, we can set up the equation as 79.5 cal/g = 200g * 1 cal/g°C * ln(T/Ti). Solving for T, we get T= 323K.

Therefore, the final temperature of the water will be 323K or 50°C. This means that the final temperature of the system will be 50°C, as the ice has melted and the energy has been dispersed throughout the system. The change in entropy of the system is the sum of the changes in entropy of the ice and water, which is 0