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.
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)
2 answers
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