Heat/fusion/vaporization question?

[pirates01]

Homework Statement

Given 150g of ice at -15^oC, how much steam (at 100^oC in g) is required to give a resultant of 50^oC water?

Homework Equations

q=lm
q=cmT
heat for vap: 2256j
heat for fusion: 333J

The Attempt at a Solution

I got something like
remaining heat: 49950-2256m
change heat with remaining water: 4.19(150+m)T
and that m(100)/(150+m)=50 (error here?)

In the end, i got something like m=9g, which is too little, cause answers says 33g

 

[kuruman]

I got something like
remaining heat: 49950-2256m

What exactly is your reasoning for this "remaining heat"? What is 49950 and how did you get it?

 

[pirates01]

remaining heat: 49950-2256m

I got this from q=lm
where 49950 is the joules taken to melt ice to liquid, and 2256m (where m is the mass of gas) is the joules given off to turn steam to liquid.

 

[kuruman]

When you end up with water at 50 ^oC, you need to say that the heat lost by the steam is equal to the heat gained by the ice. In other words,

Heat to turn steam to liquid + Heat to reduce T of condensed steam from 100 ^oC to 50 ^oC

is equal to

Heat to turn ice to liquid + Heat to raise T of melted ice from 0 ^oC to 50 ^oC.

 

[rwisz]

I would approach this problem by first identifying the known values and the unknown value. In this case, the known values are the mass of ice (150g), the initial temperature (-15oC), and the final temperature (50oC). The unknown value is the mass of steam required to reach the final temperature.

Next, I would use the equations for heat transfer to solve for the unknown value. The first equation, q=lm, represents the heat required for a substance to change phase (in this case, from ice to water). The second equation, q=cmT, represents the heat required to change the temperature of a substance.

Using these equations, I would set up two equations: one for the heat required to melt the ice and one for the heat required to raise the temperature of the resulting water. These equations would look like this:

q1 = (150g)(333J/g) = 49950J

q2 = (150g + m)(4.19J/gC)(50oC - (-15oC)) = 4.19(150g + m)(65oC) = 4.19(9750 + 65m)J

Setting these two equations equal to each other and solving for m, we get:

49950J = 4.19(9750 + 65m)J
m = 33g

Therefore, 33g of steam is required to reach a final temperature of 50oC with 150g of ice at -15oC.