Question

The heat of fusion of water is 79.9 cal/g, the heat of vaporization of water is 540 cal/g, and the specific heat of water is 1.00 cal/deg/g. How many grams of ice at 0 ° could be converted to steam at 100 °C by 9,076 cal of heat?

Answer #1

Q =heat change for conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 100 oC +heat change for conversion of water at 100 oC to vapour at 100 oC

Amount of heat absorbed , Q = mL + mc'dt + mL'

= m( L + c'dt' + L' )

Where

m = mass of ice = ?

c' = Specific heat of water = 1.00 cal/g degree C

L’ = Heat of Vaporization of water = 540 cal/g

L= Heat of fusion of water = 79.9 cal/g

dt' = 100 -0 =100 oC

Q= heat supplied= 9076 cal

Plug the values we get m = 12.6 g

Im taking a course to review general chemistry. In lecture, my
professor didnt say how to work this kind of problem in the
homework. Please list all the steps. Thanks.
The heat of fusion of water is 79.9 cal/g, the heat of
vaporization of water is 540 cal/g, and the specific heat of water
is 1.00 cal/deg/g. How many grams of ice at 0 ° could be converted
to steam at 100 °C by 9,958 cal of heat?

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