posted by luis fernandez on .
calculate the amount of energy (in kj) necessary to convert 457 g of liquid water from 0 Celecius to water vapor at 167 celcius. the molar heat of vaporization (Hvap) of water is 40.79 jk/mol. the specific heat for water is 4.187 j/g celcius and for steam 1.99 j/mol celcius. (assume that the specific heat values do not change over the range of temperature in the problem)
Note the correct spelling of celsius.
There are two equations you need to do this.
1. WITHIN a phase, q = mass x specific heat x (Tfinal-Tinitial)
2a. At the phase change, q = mass x heat fusion at melting point or
2b. At the boiling point, q = mass x heat vaporization.
For example, within a phase would be from zero C t 100 C it is liquid water.
q = mass x specific heat liquid water x (Tf-Ti). Tf = 100; Ti = 0.
So q1 = zero C to 100 C.
q2 = convert liquid water to steam at 100 C.
q3 = steam from 100 C to final T.
Then add q1 + q2 + q3.
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