At 1 atm, how much energy is required to heat 55.0 g of H2O(s) at –20.0 °C to H2O(g) at 163.0 °C?
It is best to do this in parts.
For changes in temperature WITHIN a phase, calculate q using
q = mass x specific heat x (Tfinal-Tinitial)
For example for liquid water at zero C to liquid water at 100 C. use
q = 55.0g x 4.184 J/g*C x (100-0) = ?
For q at a phaae change use
q = mass x heat fusion at melting point or
q = mass x heat vaporization at boiling point.
For esample, for water to change from solid at zero C to a liquid at zero C
Q = 55.0 x 334 J/g
Go through for 55.0 g H2O at -20 to zero, change to liquid, go from zero to 100, change to vapor, go from 100 to 163.0.
Then add all of the individual q values together.
To calculate the amount of energy required to heat a substance, you need to use the formula:
Q = m * s * ΔT
Where:
Q = energy (in joules)
m = mass of the substance (in grams)
s = specific heat capacity of the substance (in J/g °C)
ΔT = change in temperature (final temperature - initial temperature)
First, let's find the mass of water:
m = 55.0 g
Next, let's calculate the change in temperature:
ΔT = (final temperature) - (initial temperature)
= (163.0 °C) - (-20.0 °C)
= 183.0 °C
Now, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g °C.
Substituting the values into the formula:
Q = (55.0 g) * (4.18 J/g °C) * (183.0 °C)
Calculating this equation will give you the amount of energy required in joules.
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