How much heat is necessary to change a 52.0 g sample of water at 33 degrees celsius into steam at 110 degrees celsius?
q1 = heat to raise T H2O from 33 C to 100 C.
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = heat to vaporize H2O at 100 C to steam at 100 C.
q2 = mass H2O x heat vaporization.
q3 = heat to raise T of steam from 100 C to 110 C.
q3 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total q = q1 + q2 + q3.
To find the amount of heat required to change a sample of water from 33 degrees Celsius to steam at 110 degrees Celsius, we need to calculate the heat required for two processes:
1. Heating the water from 33 degrees Celsius to its boiling point.
2. Vaporizing the water at its boiling point to form steam.
Let's calculate the heat required for each step separately:
Step 1: Heating the water to its boiling point
To find the heat required for this step, we can use the specific heat capacity formula:
q1 = m × c × ΔT
where:
q1 = heat required
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
The specific heat capacity of water is approximately 4.184 J/g°C.
Using the given values:
m = 52.0 g
c = 4.184 J/g°C
ΔT = (100 - 33) = 67°C (change in temperature from 33°C to the boiling point of 100°C)
q1 = (52.0 g) × (4.184 J/g°C) × (67°C) = 14158.576 J
Step 2: Vaporizing the water to form steam
To calculate the heat required for this step, we'll use the formula:
q2 = m × ΔHvap
where:
q2 = heat required
m = mass of water
ΔHvap = enthalpy of vaporization of water
The enthalpy of vaporization of water is approximately 40.7 kJ/mol.
First, we need to convert the mass of water to moles. The molar mass of water (H2O) is 18.015 g/mol.
moles of water = (52.0 g) / (18.015 g/mol) = 2.884 moles
Then, we can calculate the heat required:
q2 = (2.884 moles) × (40.7 kJ/mol) = 117.2148 kJ = 117214.8 J
Finally, we can find the total heat required by adding q1 and q2:
Total heat required = q1 + q2 = 14158.576 J + 117214.8 J = 131373.376 J
Therefore, approximately 131,373.376 Joules of heat is necessary to change a 52.0 g sample of water at 33 degrees Celsius into steam at 110 degrees Celsius.
To calculate the heat necessary to change the temperature of water from 33 degrees Celsius to 110 degrees Celsius, we need to consider two parts: the heat required to raise the temperature of the water from 33°C to 100°C and the heat required to convert the water at 100°C into steam at 110°C.
First, let's calculate the heat required to raise the temperature of the water from 33°C to 100°C. The specific heat capacity of water is approximately 4.18 J/g·°C.
Step 1: Calculate the heat required to raise the temperature from 33°C to 100°C.
Heat = mass × specific heat capacity × temperature change
Where:
- mass = 52.0 g (the mass of water)
- specific heat capacity = 4.18 J/g·°C (specific heat capacity of water)
- temperature change = 100°C - 33°C = 67°C
Heat = 52.0 g × 4.18 J/g·°C × 67°C = 139,501.6 J (rounded to two significant figures).
Next, let's calculate the heat required for phase change from water at 100°C to steam at 110°C. The heat of vaporization for water is approximately 2260 J/g.
Step 2: Calculate the heat required for phase change.
Heat = mass × heat of vaporization
Where:
- mass = 52.0 g (the mass of water)
- heat of vaporization = 2260 J/g (heat of vaporization for water)
Heat = 52.0 g × 2260 J/g = 117,520 J (rounded to two significant figures).
Finally, let's calculate the total heat required to change the water to steam at 110°C.
Total Heat = Heat for temperature change + Heat for phase change
= 139,501.6 J + 117,520 J
= 257,021.6 J (rounded to two significant figures).
Therefore, approximately 257,021.6 J (or 2.57 × 10^5 J) of heat is required to change a 52.0 g sample of water at 33 degrees Celsius into steam at 110 degrees Celsius.