Calculate Change in Entropy of Ice Cube-Water System

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In summary, the problem involves an ice cube of 8.5 g at -10°C being placed into a Thermos flask containing 200 cm3 of water at 20°C. The specific heat of ice is 2220 J/kg·K. The goal is to find the change in entropy of the cube-water system at equilibrium. Using the equations given, the final temperature of the system is calculated to be 288.73 K. The change in entropy is then found to be 6.67 J/K, taking into account the melting of the ice, the change in temperature of the ice, and the change in temperature of the water.

Homework Statement

An 8.5 g ice cube at -10°C is put into a
of water at 20°C. By how much has the
entropy of the cube-water system changed when equilibrium
is reached? The specific heat of ice is 2220 J/kg·K.

Cw= 4190
Mi= 8.5
Mw= 200
Lf= 333000
Ci=2220

Homework Equations

Tf=cw*mw(20)-mi(lf+10ci)/mw*cw+mi*cw

Tf= 288.73 K

mi*ci*ln(273.15/263.15)+lf*mi/273.15+mi*cw*ln(288.73/273.15)+mw*cw*ln(288.73/293.15)

The Attempt at a Solution

.70366+10.36+1.975-6.365

I got 6.67 J/k. But I'm not sure that's right.

It is not very clear what you are trying to do here.

The first thing you have to do is figure out whether all the ice melts. If it does, you then have to figure out what the final temperature is.

Then you have to work out what the change in entropy of the ice is in reaching 0C, in changing to water, and in rising to the final temperature. Use the definition of incremental change in entropy: dS = dQ/T.

Then you have to work out the change in entropy of the 200 g of water in cooling to the final temperature.

AM

1. How do you calculate the change in entropy of an ice cube-water system?

To calculate the change in entropy of an ice cube-water system, you need to know the initial and final temperatures of the system, the specific heat capacity of water, and the latent heat of fusion of water. The equation for calculating entropy change is ΔS = Q/T, where Q is the heat added or removed from the system and T is the temperature in Kelvin. The latent heat of fusion of water is used to account for the change in state from solid to liquid.

2. What factors affect the change in entropy of an ice cube-water system?

The change in entropy of an ice cube-water system is affected by temperature, heat added or removed, and the change in state of water molecules. Higher temperatures and more heat added will result in a larger change in entropy, while lower temperatures and less heat added will result in a smaller change in entropy. Additionally, the change in state from solid to liquid also contributes to the change in entropy.

3. Why is the change in entropy of an ice cube-water system important to understand?

The change in entropy of an ice cube-water system is important to understand because it is a fundamental concept in thermodynamics. It helps us understand how energy is transferred and transformed in a system, and how the disorder or randomness of the system changes. It also has practical applications, such as in refrigeration and phase changes.

4. Can the change in entropy of an ice cube-water system be negative?

Yes, the change in entropy of an ice cube-water system can be negative. This would occur if the heat removed from the system is greater than the heat added, resulting in a decrease in disorder or randomness. This is commonly seen in the process of freezing, where the water molecules become more organized as they transition from liquid to solid state.

5. How does the change in entropy of an ice cube-water system relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of an isolated system will always increase over time. In the case of an ice cube-water system, the change in entropy will depend on the initial and final temperatures, as well as the heat added or removed. The second law of thermodynamics can be applied to understand the direction of heat flow and the change in disorder or randomness in the system.