Calorimetry Help: Solving a Problem with Ice Phase Change

[PhysicsinCalifornia]

I needed to know if anyone can help me solve a problem.

I am supposed to calculate the final temperature after mixing 250 mL of coffee at 80 degrees C with 250 g of ice at 0 degrees C.

The heat of fusion for ice is 6.01 kJ/mol

The ice undergoes a phase change, but I don't know how to apply it. Can anyone help me?

 

[dav2008]

Well the heat of fusion says it all. It is 6.01 kJ/mol. That means that it takes 6.01 kJ of heat to melt 1 mol of ice--that is changing 1 mol of ice at 0 degrees C to 1 mol of water at 0 degrees C.

Technically you're supposed to show some work before you can get help here, so if you need more help please post to what extent you've worked the problem out so far.

 

[Blueshift5]

Sure, I can definitely help you with this problem! First, we need to understand the principles of calorimetry. Calorimetry is the measurement of heat transfer in a system. In this case, we have a system consisting of hot coffee and ice, and we want to calculate the final temperature after they are mixed together.

To solve this problem, we will use the equation Q = mCΔT, where Q is the heat transferred, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.

First, we need to determine the heat released by the coffee (Qcoffee) and the heat absorbed by the ice (Qice). Qcoffee can be calculated by multiplying the mass of the coffee (250 mL) by its specific heat capacity (4.18 J/g°C) and the change in temperature (80°C - Tf, where Tf is the final temperature). Qice can be calculated by multiplying the mass of the ice (250 g) by the heat of fusion (6.01 kJ/mol).

Next, we can set up an equation to represent the conservation of energy in this system: Qcoffee = -Qice. This means that the heat released by the coffee is equal to the heat absorbed by the ice.

Now we can substitute the values we have into the equation: (250 mL)(4.18 J/g°C)(80°C - Tf) = -(250 g)(6.01 kJ/mol).

Solving for Tf, we get a final temperature of 6.08°C. This is the temperature at which the coffee and ice will reach thermal equilibrium.

It's important to note that we did not take into account the heat capacity of the container or any heat loss to the surroundings. These factors may affect the final temperature slightly, but for the purposes of this problem, we can assume they are negligible.

I hope this helps you understand how to solve calorimetry problems involving phase changes. Remember to always pay attention to units and use the appropriate values for specific heat capacities and heat of fusion. Let me know if you have any further questions!