Two kg of water at a temperature of 50 degrees C have 485.3 kJ of heat transferred from them, resulting in a mixture of ice/water at 0 degrees C. What is the composition of the mixture?
This is what I did and would like verification of my answer.
2*4.183*-50 = -418.3 kJ
I then subtracted that from my total heat energy:
485.3 - 418.3 = 67 kJ
For latent fusion:
67 = 335*m
m = 0.2 kg of ice
My original mass of water was 2 kg so the composition of the final mixture would be:
0.2 kg of ice
1.8 kg of water
Your calculations are correct, and the composition of the mixture is:
- 0.2 kg of ice
- 1.8 kg of water
To further explain the process:
1. First, you calculated the heat lost by the water at 50 degrees C using the equation Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Substituting the values, you correctly obtained -418.3 kJ.
2. Next, you subtracted the heat lost from the total heat transferred to find the heat energy absorbed by the ice/water mixture. Subtracting 418.3 kJ from 485.3 kJ gave you 67 kJ.
3. To find the mass of the ice, you used the equation Q = mLf, where Q is the heat energy, m is the mass of the ice, and Lf is the latent heat of fusion. Substituting the values, you correctly obtained a mass of 0.2 kg.
4. Since the original mass of water was 2 kg, you could calculate the remaining mass of water in the mixture by subtracting the mass of ice from the original mass of water, which gave you 1.8 kg.
Therefore, the composition of the mixture is 0.2 kg of ice and 1.8 kg of water.