Find Ice Melted & Final Temp w/ 10g Steam & 50g Ice

[Jennifer001]

Homework Statement

steam at 100C is added to ice at 0C find the amount of ice melted and the final temperature when the mass of steam is 10g and the mass of ice is 50g

Homework Equations

Q=mcdT
Q=ML

The Attempt at a Solution

Qmelt=MiciT+MiLf
=0.05(2090)(100C)+(0.05)(3.33x10^5)
=27100J

27100+Micw(Tf-0C)+Mscw(Tf-20C)=0

Tf=[100(0.01)(4190)-27100]/[(0.05)(4190)+(0.01)(4190)]
=-91

am i doing this question correctly? cause in the question it askes for the final temperature but i don't think the finall temperture would be -91 when it starts out with 100C and 0C.. also I am not sure how to find the amount of ice melted can someone help?

i thnk it has something to do with finding out the final temperture then plug it into Q=mcdT and you solve for m?? is this correct?.. but i still don't know how to find the final temperture correctly.. please help

 

[mgb_phys]

Since energy is conserved you have:
Energy lost by steam changing to water + Energy lost by that water cooling to T = Energy needed to melt ice + energy needed to heat that water to T

 

[Jennifer001]

so is the equation:

MscwdT+MwcwdT=MicidT+MwcwdT

Ms= mass of steam
Mw= mass of water
Mi= mass of ice..
sorry I'm really lost in thermodynamics

 

[mgb_phys]

Not quite.
The heat of melting/boiling doesn't have a temperature - there is no T change.
When the steam condenses it is cooling from 100deg, this water loses energy so it's temperature change is (100-T)

You will also need these values (or whatever is given in the question)
Specific heat water, c = 4.184 J/g.K
Heat vapourisation = 2260 J/g
Heat fusion = 333.5 J/g

 

[fragglerock11]

I'm pretty sure you're going to need to incorporate the latent heats of fusion in there.

and mgb_phys, are you sure it's (100-T) Shouldn't the change in temperature always be Tf-Ti?

unless you're going to throw the magnitude brackets around it?

 

[mgb_phys]

T is the final temperature, the question needs \deltaT, which for the steam is (100-
T) and for the ice (0+T).