# Solve Entropy Problems: Find Energy Change & Entropy Change

• salient
In summary: I was trying to solve this the other day and I got so far before I crashed and burned. I was using the equation for entropy change, but I got confused about what temperature I should be solving for. Turns out that you have to solve for the temperature at which the entropy change is equal to the Q change, not the final temperature. So if I were to solve for the temperature at which the entropy change is equal to the Q change, it would be Tf-0.
salient

## Homework Statement

80g of ice initially at zero degrees C is allowed to melt in 200g of water initially at 20 degrees C.

Find the energy change of the water, energy change of ice
Find entropy change in water, ice, and universe.

## Homework Equations

Delta S = Q/T
Q = (Specific heat)(mass)(Delta T)
Lf = heat of fusion
Cw = specific heat of water (yes I know it's 4.184ish J/grams degrees C, but he doesn't like us using numbers).

## The Attempt at a Solution

My physics instructor spent a half-hour defining entropy in about 18 different unique-to-him ways, and an hour and a half on random tangents that didn't really have much to do with entropy or even physics. No equations were given, no problems were demonstrated.

So I'm just going to attempt to solve this using methods I remember from high school chemistry about fifteen years ago, and I'm hoping maybe someone can tell me if I'm way off my rocker.

so Qice = (80)(Lf) + (80)(Cw)(Tf-0)
and Qwater = (200)(Cw)(Tf-20)

Since I know energy from the water is going into the ice:

(80)(Lf) + (80)(Cw)(Tf-0) = -(200)(Cw)(Tf-20)

Since the final temperature is the only real variable there, I can solve for that and come up with values for the Qs.

As far as entropy goes though, this isn't a constant temperature process. In the equation Delta S = Q/T, if indeed that is the equation I should be using, do I put the final temperature there below the Q change required to reach it? Would the entropy change of the universe be the sum of those two Delta Ss?

I suppose that this whole process is adiabatic, there is no heat exchange with the environment, and there is no work done, or it can be ignored.
So you have three processes: Melting ice is isotherm, the temperature of ice stays 0 C°till it completely melts. You can get the entropy change during melting from the heat and temperature.
For the entropy change of the water, take into account that dQ=cmdT, so dS=dQ/T=cmdT/T. Integrate between the initial and final temperature to get the change of entropy. Do the same with both the warm water and with the melted ice.
The environment does not get any heat if the process was adiabatic, so its entropy does not change. So the entropy change of the universe is the sum of the three entropy terms: melting the ice, warming up the ice-water and cooling down the warm water.

ehild

salient said:
As far as entropy goes though, this isn't a constant temperature process. In the equation Delta S = Q/T, if indeed that is the equation I should be using, do I put the final temperature there below the Q change required to reach it? Would the entropy change of the universe be the sum of those two Delta Ss?
Yes.
Entropy is calculated along the reversible path between initial and final states.

$$\Delta S_{universe} = \Delta S_{ice} + \Delta S_{water} = \int_{273}^{T_f} dQ_{rev}/T + \int_{293}^{T_f} dQ_{rev}/T = \int_{273}^{T_f} (m_{ice}Lf +m_{ice}CdT) /T + \int_{293}^{T_f} m_{water}CdT/T$$AM

## 1. What is entropy and why is it important?

Entropy is a measure of the disorder or randomness in a system. It is important because it helps us understand the direction and extent of chemical and physical processes, and can be used to predict the spontaneity of reactions.

## 2. How do you calculate the change in entropy?

The change in entropy (ΔS) can be calculated by taking the sum of the products of the molar entropy of each reactant/ product and the corresponding stoichiometric coefficient, and subtracting it from the sum of the same calculation for the products. In equation form, it is written as: ΔS = ∑(nS(products)) - ∑(nS(reactants)), where n is the stoichiometric coefficient and S is the molar entropy.

## 3. How do you determine the energy change in a system using entropy?

The energy change (ΔH) in a system can be determined by using the equation: ΔH = TΔS, where T is the temperature in Kelvin. This equation is derived from the relationship between entropy and the second law of thermodynamics, which states that in any spontaneous process, the total entropy of the universe must increase.

## 4. Can entropy be negative?

In most cases, entropy cannot be negative. However, in some limited cases, such as when a gas is compressed or when a pure crystalline solid is formed from a solution, the entropy change can be negative. This is because these processes result in a decrease in randomness and disorder within the system.

## 5. How does entropy relate to the concept of equilibrium?

Entropy plays a crucial role in determining the direction of chemical reactions and the attainment of equilibrium. The second law of thermodynamics states that in any spontaneous process, the total entropy of the universe must increase. Therefore, in order to reach equilibrium, a system will tend to move towards the state with the highest possible entropy, where the system is in a state of maximum disorder.

• Introductory Physics Homework Help
Replies
5
Views
240
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
857
• Introductory Physics Homework Help
Replies
2
Views
802
• Introductory Physics Homework Help
Replies
2
Views
998
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
918