Thermal Equilibrium and Energy System/Phase Diagrams

[elissa5]

Homework Statement

A 0.5kg piece of ice(water) at an initial temp of -65 degrees Celsius is placed inside a well-insulated container with 3.0 kg of tea at an initial temp of 20 degrees Celsius, and the two are allowed to come to thermal equilibrium( the tea can be treated as water with respect to thermal properties)... they want me to draw a energy system diagram to show this using ethermal and Ebond and figure out the thermal equilibrium?

Homework Equations

Ethermal= mcp(tf-ti)
Ebond= delta(m)delta(h)
mass of ice= 0.5kg
mass of water= 3.0kg
heat of ice/water/watervapor= 333.5 kJ/kg
specific heat of ice 2.05 KJ/kgK
specific heat of water 4.18 kJ/kgK

The Attempt at a Solution

1. The part I am mainly confused about is i know we have to figure out delta(m)delta(h) for ebond and i have the solution saying that delta m is .5 and delta h is 333.5= 166.7. I understand how they got delta h but I am unsure of how they got deltam to be .5?? Another thing is i know that we have to do mc(tf-ti) for tea which is (3)(4.18)(0+20)=250.8 and we do the same thing for ice which is (.5)(2.05)(65)=66. But what's the point of figuring out all these numbers...how do we use these to solve the problem (it has to do with phase diagrams but not sure how)??

Sorry i know this problem is kinda long. But any help would be greatly appreciated! Thanks! The solution from the book states Te= 1.19.

 

[jbsteele]

Tf= -14.3 The way to approach this problem is through the energy balance. You have to think of it as an energy exchange between the ice and the tea. The energy balance equation is Ethermal + Ebond = 0. The thermal energy is given by mCp(Tf - Ti). For the water, that would be 3(4.18)(Tf - 20) and for the ice, it would be 0.5(2.05)(Tf + 65). The bond energy is given by ΔMΔH, where ΔM is the difference in mass between the two materials and ΔH is the heat of fusion (or vaporization) of either water or ice. In this case, ΔM is 0.5 kg and ΔH is 333.5 kJ/kg. So the bond energy is 0.5(333.5) = 166.7 kJ. Now, you can solve the energy balance equation for Tf, the temperature of the equilibrium state. Note that the thermal energy of the ice is negative because its temperature is below 0°C. So the overall equation is 3(4.18)(Tf - 20) + 0.5(2.05)(Tf + 65) - 166.7 = 0. Solving this equation yields Tf = -14.3°C. Since the temperatures of both the ice and tea are now equal, they are in equilibrium.

 

[kerimek]

First of all, in order to understand the solution, it is important to have a basic understanding of thermal equilibrium and energy systems. Thermal equilibrium is a state where two objects in contact with each other reach the same temperature and there is no net heat transfer between them. In this case, the ice and tea will reach the same temperature at thermal equilibrium.

To solve this problem, we need to consider the total energy of the system. The energy of the system can be divided into two parts: Ethermal and Ebond. Ethermal is the thermal energy of the system, which is related to the temperature of the system. Ebond is the energy required to change the state of the substance, in this case, from ice to water.

First, let's calculate the Ethermal for the tea and ice separately. For the tea, we use the equation Ethermal = mcp(tf-ti), where m is the mass, c is the specific heat, and tf and ti are the final and initial temperatures, respectively. We know the mass of the tea is 3.0 kg and the specific heat of water is 4.18 kJ/kgK. The initial temperature of the tea is 20 degrees Celsius, and the final temperature is the same as the final temperature of the ice, which we will calculate later. Therefore, Ethermal of the tea is (3)(4.18)(Tf-20).

For the ice, we use the same equation, but we have to consider the change in temperature from -65 degrees Celsius to the final temperature Tf. Therefore, Ethermal of the ice is (0.5)(2.05)(Tf-(-65)).

Next, we need to calculate the Ebond for the ice. Ebond is the energy required to melt the ice, which is related to the change in mass and the heat of fusion. We know the heat of fusion for ice is 333.5 kJ/kg. The change in mass is simply the mass of the ice, which is 0.5 kg. Therefore, Ebond for the ice is (0.5)(333.5).

Now, we can equate the Ethermal and Ebond for the tea and ice and solve for Tf. This is because at thermal equilibrium, the total energy of the system must be the same. Therefore, we have the equation (3)(4.18)(Tf-20) = (0.5)(2.05