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