Calculater the amount of heat in kilojoules released when 50 grams of steam at 125 C are converted to 50 grams of ice at -125 C?
q1 = heat released going from steam at 125 to steam at 100.
q1 = mass steam x specific heat steam x delta T.
q2 = heat released to condense steam at 100 to liquid water at 100.
q2 = mass steam x heat vaporization.
q3 = heat released going from water at 100 to water at zero.
q3 = mass H2O x specific heat water x delta T.
q4 = heat released to freeze water at zero to ice at zero.
q4 = mass water x heat of fusion.
q5 = heat released for ice going from zero to -125.
q5 = mass ice x specific heat ice x delta T.
Total q = q1 + q2 + q3 + q4 + q5.
Post your work if you get stuck.
To calculate the amount of heat released during a phase change, we can use the equation:
Q = m × H
Where:
Q is the amount of heat released or absorbed in joules,
m is the mass of the substance in kilograms, and
H is the specific heat capacity of the substance in joules per kilogram per degree Celsius.
Here's how we can solve the problem step by step:
Step 1: Convert the temperature units to Kelvin.
To convert Celsius to Kelvin, we add 273.15 to the temperature.
125°C + 273.15 = 398.15 K
-125°C + 273.15 = 148.15 K
Step 2: Determine the amount of heat released during the phase change from steam to water at 100°C.
We know that the specific heat capacity for vaporization (latent heat of vaporization) of water is 2260 J/g. However, the units are given in grams, so we must convert grams to kilograms:
50 grams = 0.05 kilograms
Q1 = m × Hvap
= 0.05 kg × 2260 J/g
= 113 J
Step 3: Determine the amount of heat released during the cooling process from 100°C to 0°C.
We can use the specific heat capacity (SHC) of water for this calculation, which is approximately 4.18 J/g°C.
Q2 = m × SHC × ∆T
= 0.05 kg × 4.18 J/g°C × (100°C - 0°C)
= 20.9 J
Step 4: Determine the amount of heat released during the phase change from water to ice at 0°C.
The specific heat capacity for freezing (latent heat of fusion) of water is 334 J/g.
Q3 = m × Hfus
= 0.05 kg × 334 J/g
= 16.7 J
Step 5: Determine the amount of heat released during the cooling process from 0°C to -125°C.
Using the specific heat capacity (SHC) of ice, which is approximately 2.09 J/g°C, we can calculate:
Q4 = m × SHC × ∆T
= 0.05 kg × 2.09 J/g°C × (0°C - (-125°C))
= 16.12 J
Step 6: Sum up the heat released during each step to obtain the total heat released.
Total heat released = Q1 + Q2 + Q3 + Q4
= 113 J + 20.9 J + 16.7 J + 16.12 J
≈ 166.72 J
Step 7: Convert the heat from joules to kilojoules.
To convert from joules (J) to kilojoules (kJ), divide the value by 1000:
Total heat released (kJ) = 166.72 J / 1000
≈ 0.1667 kJ
Therefore, the amount of heat released when 50 grams of steam at 125°C are converted to 50 grams of ice at -125°C is approximately 0.1667 kJ.