How many kilojoules of heat is required to completely covert 60.0 grams of water at 32.0° C to steam at 100°C?
For water:
s = 4.179 J/g °C
Hfusion = 6.01 kJ/mol
Hvap = 40.7 kJ/mol
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To find the amount of heat required to convert the given amount of water from 32.0°C to steam at 100°C, we need to calculate the heat required for each step of the process:
Step 1: Heating the water from 32.0°C to 100°C
Step 2: Vaporizing the water at 100°C
Step 1: Heating the water from 32.0°C to 100°C
To calculate the amount of heat required to heat the water, we use the formula:
q = m * Cp * ΔT
where:
q is the amount of heat (in joules),
m is the mass of the substance (in grams),
Cp is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
Given:
m = 60.0 grams
Cp (specific heat capacity of water) = 4.179 J/g °C
ΔT = (100°C - 32.0°C) = 68.0°C
Using the formula, we can calculate the amount of heat required to heat the water:
q1 = m * Cp * ΔT
= 60.0 g * 4.179 J/g °C * 68.0°C
= 16,068.24 J
Step 2: Vaporizing the water at 100°C
To calculate the amount of heat required to vaporize the water, we use the formula:
q = m * Hvap
where:
q is the amount of heat (in joules),
m is the mass of the substance (in grams), and
Hvap is the enthalpy of vaporization (in J/g).
Given:
m = 60.0 grams
Hvap (enthalpy of vaporization) = 40.7 kJ/mol = 40.7 kJ/18.015 g (since 1 mole of water is 18.015 g)
Using the formula, we can calculate the amount of heat required to vaporize the water:
q2 = m * Hvap
= 60.0 g * (40.7 kJ/18.015 g)
= 134.0327 kJ
To find the total heat required, we sum up the heat required for each step:
Total heat = q1 + q2
= 16,068.24 J + 134.0327 kJ
= 150,100.94 J
Therefore, approximately 150,100.94 Joules of heat is required to completely convert 60.0 grams of water at 32.0°C to steam at 100°C.