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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?
2 answers
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.
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.
1 answer