Hi, I just need to know if I finally got this exercise right.
Calculate the amount of heat in kilojoules released when 50 grams of steam at 125 Celsius are converted to 50 grams of ice at -125 Celsius.
so,
qw= 50*4.184J/g C (125 C-o)= 26.15 x 10^3
qi= 50*2.09*(-125)=-13.065 KJ
q= 50*1.860 (-125)= -11.63 KJ
q1+q2=q3= -11.92 kJ
please let me know if this way is correct thanks
No, I didn't read past the first line. Steam is not liquid water; therefore, you may not use the specific heat of water. You must use the specific heat of steam. THEN, it doesn't go from 125 to zero. It goes from 125 to 100.
In the initial post I told you what we were converting; i.e., from steam at 125 to steam at 100, then condense the vapor steam to liquid water at 100, then move form liquid water at 100 to liquid water at zero, then heat of fusion to freeze water to ice at zero, then move from ice at zero to ice at -125.
To determine if your calculations are correct, let's break down the steps and confirm each calculation.
1. Start by calculating the heat released when 50 grams of steam at 125 degrees Celsius are converted to water at 100 degrees Celsius. This step involves the heat released during the cooling process:
q1 = mass * specific heat capacity * change in temperature
Given:
- mass = 50 grams
- specific heat capacity of water/steam = 4.184 J/g°C
- change in temperature = (125°C - 100°C) = 25°C
Calculating q1:
q1 = 50 g * 4.184 J/g°C * 25°C = 5,230 J
Note: Make sure to multiply the mass by the specific heat capacity and the change in temperature to get the correct unit of energy (Joules in this case).
2. Next, calculate the heat released during the phase change from water to ice:
q2 = mass * heat of fusion
Given:
- mass = 50 grams
- heat of fusion of water = 334 J/g
Calculating q2:
q2 = 50 g * 334 J/g = 16,700 J
3. Finally, calculate the heat released when the water freezes and cools from 0 degrees Celsius to -125 degrees Celsius:
q3 = mass * specific heat capacity * change in temperature
Given:
- mass = 50 grams
- specific heat capacity of ice = 2.09 J/g°C
- change in temperature = (0°C - (-125°C)) = 125°C
Calculating q3:
q3 = 50 g * 2.09 J/g°C * 125°C = 13,063 J
Now, let's convert the values to kilojoules (kJ):
q1 = 5,230 J = 5.23 kJ
q2 = 16,700 J = 16.70 kJ
q3 = 13,063 J = 13.06 kJ
To calculate the total heat released during the entire process, we add q1, q2, and q3:
q_total = q1 + q2 + q3 = 5.23 kJ + 16.70 kJ + 13.06 kJ = 35.99 kJ
So, the correct answer is approximately 35.99 kJ. It appears that your calculation for q3 is incorrect.