22.) The molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of liquid water is 4.18 J/gC. (1) What quantity of heat is required to melt 25.0 g of ice at 0C? (2) What quantity of heat is required to vaporoze 3.75 g of liquid water at 100C? (3) What quantity of heat is required to heat to warm 55.2 g of liquid water from 0C to 100C
(1) 8350 J
(2) 84.6 J
(3) 23.1 J
23.) Given that the specific heat capacities of ice and steam are 2.06 J/gC and 2.03 J/gC, respectively, and considering the information about water given in Problem 22, calculate the total quantity of heat evolved when 10.0 g of steam at 200C is condensed, cooled, and frozen to ice at -50C.
1. q = mass ice x heat fusion
2. q = mass H2O x heat vaporization
3. q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
23. A Loooong problem. Here is the way you do it.
Formula for q WITHIN a phase.
mass x specific heat x (Tfinal-Tinitial).
Formula for q at a phase change.'
freezing: mass x heat fusion
boiling: mass x hat fusion.
Then add the segments for total q.
Oh, the things we do for ice! Let's calculate the quantity of heat evolved when steam becomes ice.
First, we need to find the amount of heat required to cool the steam from 200C to 100C. To do that, we multiply the mass of the steam (10.0 g) by the specific heat capacity of steam (2.03 J/gC) and then by the temperature change (200C - 100C = 100C). So, the heat required for this step is 10.0 g * 2.03 J/gC * 100C = 2030 J.
Next, we need to consider the heat required for the phase change, which is when the steam condenses into liquid water. We can use the molar heat of vaporization (40.6 kJ/mol) to find the heat required. Since the molar mass of water is approximately 18 g/mol, the heat required for this step is (40.6 kJ/mol) / (18 g/mol) = 2255.6 J/g. Multiplying this by the mass of the steam (10.0 g), we get 2255.6 J/g * 10.0 g = 22,556 J.
Finally, we need to cool the liquid water to its freezing point, which is 0C. Using the specific heat capacity of water (4.18 J/gC), we can calculate the heat required. The temperature change is 100C, and the mass of the water is 10.0 g, so the heat required is 4.18 J/gC * 100C * 10.0 g = 4180 J.
Now, let's add up all the heats required: 2030 J + 22,556 J + 4180 J = 27,766 J.
So, the total quantity of heat evolved when 10.0 g of steam at 200C is condensed, cooled, and frozen to ice at -50C is 27,766 J. That's quite a frosty adventure!
Please note that the temperatures expressed in Celsius should be converted to Kelvin for more accurate calculations.
To solve this problem, we need to calculate the total quantity of heat evolved when steam is condensed, cooled, and frozen to ice. We can break down the process into three steps:
Step 1: Heat evolved during condensation from steam to water at 100°C.
Step 2: Heat evolved during cooling of water from 100°C to 0°C.
Step 3: Heat evolved during freezing of water to ice at 0°C.
Step 1: To calculate the heat evolved during condensation, we can use the molar heat of vaporization for water, which is 40.6 kJ/mol. We need to calculate the number of moles of water vapor condensed. The molecular weight of water is 18 g/mol.
Mass of steam = 10.0 g
Number of moles = mass/molecular weight = 10.0 g / 18 g/mol ≈ 0.556 mol
Heat evolved during condensation = number of moles × molar heat of vaporization
= 0.556 mol × 40.6 kJ/mol = 22.59 kJ
= 22.59 kJ × 1000 J/kJ = 22590 J
Step 2: To calculate the heat evolved during cooling, we need to use the specific heat capacity of water, which is 4.18 J/g°C. We need to calculate the temperature change from 100°C to 0°C.
Mass of water = 10.0 g
Temperature change = 100°C - 0°C = 100°C
Heat evolved during cooling = mass × specific heat capacity × temperature change
= 10.0 g × 4.18 J/g°C × 100°C = 4180 J
Step 3: To calculate the heat evolved during freezing, we can use the molar heat of fusion for water, which is 6.02 kJ/mol. We need to calculate the number of moles of water frozen.
Mass of ice = 10.0 g
Number of moles = mass/molecular weight = 10.0 g / 18 g/mol ≈ 0.556 mol
Heat evolved during freezing = number of moles × molar heat of fusion
= 0.556 mol × 6.02 kJ/mol = 3.36 kJ
= 3.36 kJ × 1000 J/kJ = 3360 J
Total heat evolved = Heat evolved during condensation + Heat evolved during cooling + Heat evolved during freezing
= 22590 J + 4180 J + 3360 J
= 30330 J
Therefore, the total quantity of heat evolved when 10.0 g of steam at 200°C is condensed, cooled, and frozen to ice at -50°C is approximately 30330 J.
To answer these questions, we need to apply the principles of heat transfer and phase changes. To calculate the heat required or released, we can use the following formulas:
1. Heat to melt/freeze: q = mass × heat of fusion
2. Heat to vaporize/condense: q = mass × heat of vaporization
3. Heat transfer within a phase: q = mass × specific heat capacity × change in temperature
Let's solve each question step by step:
(1) What quantity of heat is required to melt 25.0 g of ice at 0°C?
To melt the ice, we need to calculate the heat required for the phase change (from solid to liquid) and the heat required for heating the resulting liquid water.
1. Heat to melt: q1 = 25.0 g × 6.02 kJ/mol / (18.02 g/mol) = 8.34 kJ
2. Heat to warm up: q2 = 25.0 g × 4.18 J/g°C × (0 - (-0))°C = 0 J (no temperature change)
The total heat required is: q_total = q1 + q2 = 8.34 kJ + 0 J = 8.34 kJ = 8340 J
So, the quantity of heat required to melt 25.0 g of ice at 0°C is 8340 J.
(2) What quantity of heat is required to vaporize 3.75 g of liquid water at 100°C?
To vaporize the liquid water, we need to calculate the heat required for the phase change (from liquid to gas) and the heat required for heating the resulting steam.
1. Heat to vaporize: q1 = 3.75 g × 40.6 kJ/mol / (18.02 g/mol) = 8.34 kJ
2. Heat to warm up: q2 = 3.75 g × 4.18 J/g°C × (100 - 100)°C = 0 J (no temperature change)
The total heat required is: q_total = q1 + q2 = 8.34 kJ + 0 J = 8.34 kJ = 8340 J
So, the quantity of heat required to vaporize 3.75 g of liquid water at 100°C is 8340 J.
(3) What quantity of heat is required to warm 55.2 g of liquid water from 0°C to 100°C?
To warm up the liquid water, we only need to calculate the heat required for a temperature change within the same phase.
Heat to warm up: q = 55.2 g × 4.18 J/g°C × (100 - 0)°C = 23,089.6 J ≈ 23.1 kJ
So, the quantity of heat required to warm up 55.2 g of liquid water from 0°C to 100°C is approximately 23.1 kJ.
Now, let's move on to question 23:
Given the specific heat capacities of ice and steam, we can calculate the heat transferred during each phase change and temperature change from steam at 200°C to ice at -50°C.
The total quantity of heat evolved is obtained by adding the heat released during each step.
1. Heat to cool the steam to 100°C:
q1 = 10.0 g × 2.03 J/g°C × (100 - 200)°C = -2,030 J = -2.03 kJ (negative because heat is released)
2. Heat to condense the steam to liquid water:
q2 = 10.0 g × 40.6 kJ/mol / (18.02 g/mol) = 22.6 kJ
3. Heat to cool the liquid water to 0°C:
q3 = 10.0 g × 4.18 J/g°C × (0 - 100)°C = -4,180 J = -4.18 kJ (negative because heat is released)
4. Heat to freeze the liquid water to ice:
q4 = 10.0 g × 2.06 J/g°C × (-50 - 0)°C = -10,300 J = -10.3 kJ (negative because heat is released)
The total quantity of heat evolved is: q_total = q1 + q2 + q3 + q4 = -2.03 kJ + 22.6 kJ - 4.18 kJ - 10.3 kJ = 6.09 kJ = 6,090 J.
So, the total quantity of heat evolved when 10.0 g of steam at 200°C is condensed, cooled, and frozen to ice at -50°C is 6,090 J.