Question

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?

Answer 1

Q = m c ∆T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g ^oC)
∆T = change in temperature = Tfinal - Tinitial in ^oC

specific heat (c) = Heat Energy / (mass of substance * change in temperature)

The "cold water" in this case is going to do three things:

a) icewill melt
b) as a liquid, the temperature goes up 100 °C

c) Liquid has to convert to steam

Let us keep w is weight of water .

w x 79.9 /18 + w x 1 x 100 + w x 540 /18 = 9,958

4.438 w + 100 w + 30 w = 9958

134.438 w = 9958

w = 74.07 gm

Weight of water is 74.07 gm