How much heat in kilojoules is evolved in converting 1.50moles of steam at 150 degrees celcius to ice at -60.0 degrees celcius? The heat capacity of steam is 1.84J/g degree C and that of ice is 2.09J/g degree C.
To calculate the heat evolved in converting steam to ice, we need to consider the following steps:
Step 1: Calculate the heat required to cool steam from 150 °C to 0 °C.
Step 2: Calculate the heat required to convert steam at 0 °C to water at 0 °C.
Step 3: Calculate the heat required to cool water from 0 °C to -60 °C.
Step 4: Calculate the heat required to freeze water at -60 °C to ice at -60 °C.
Step 5: Sum up the heats calculated in steps 1-4 to get the total heat evolved.
Let's begin with step 1:
The heat capacity of steam is given as 1.84 J/g°C. To calculate the heat required to cool 1.50 moles of steam from 150 °C to 0 °C, we need the molar mass of water, which is approximately 18.015 g/mol.
Using the equation: q = m × C × ΔT, where q is the heat, m is the mass in grams, C is the specific heat capacity in J/g°C, and ΔT is the change in temperature in °C.
q1 = (mass of steam) × (specific heat capacity of steam) × (change in temperature)
q1 = (1.50 moles × 18.015 g/mol) × (1.84 J/g°C) × (0 - 150 °C)
Now, solve for q1 to get the heat required to cool steam from 150 °C to 0 °C.
Next, move on to step 2:
To convert steam at 0 °C to water at 0 °C, we need to calculate the heat of fusion. The heat of fusion of water is 334 J/g.
Using the equation: q = m × ΔHf, where q is the heat, m is the mass in grams, and ΔHf is the heat of fusion in J/g.
q2 = (mass of steam) × (heat of fusion of water)
q2 = (1.50 moles × 18.015 g/mol) × (334 J/g)
Now, solve for q2 to get the heat required to convert steam to water.
For step 3:
We need to calculate the heat required to cool water from 0 °C to -60 °C. The heat capacity of water is equal to the heat capacity of ice, which is given as 2.09 J/g°C.
Using the equation: q = m × C × ΔT, where q is the heat, m is the mass in grams, C is the specific heat capacity in J/g°C, and ΔT is the change in temperature in °C.
q3 = (mass of water) × (specific heat capacity of water) × (change in temperature)
q3 = (1.50 moles × 18.015 g/mol) × (2.09 J/g°C) × (0 - (-60) °C)
Now, solve for q3 to get the heat required to cool water from 0 °C to -60 °C.
Finally, for step 4:
To freeze water at -60 °C to ice at -60 °C, we need to calculate the heat of fusion of water at -60 °C. The heat of fusion remains the same at 334 J/g.
Using the equation: q = m × ΔHf, where q is the heat, m is the mass in grams, and ΔHf is the heat of fusion in J/g.
q4 = (mass of water) × (heat of fusion of water)
q4 = (1.50 moles × 18.015 g/mol) × (334 J/g)
Now, solve for q4 to get the heat required to freeze water at -60 °C to ice at -60 °C.
Finally, for step 5:
Sum up the heats calculated in steps 1-4:
Total heat evolved = q1 + q2 + q3 + q4
Now you can plug in the values calculated in each step and calculate the final result.
break up the question into parts, then add them
cooling steam to 100c
condensing steam at100
cooling water from 100C to OC
converting to ice at OC