Physics Heat Ice Water Mixture Question And Find Initial Mass Of Ice

[justinh8]

Homework Statement

An ice-water mixture has a mass of 180g and is contained in a 100g aluminum calorimeter When 35g of steam at 100 degree Celsius is condensed in the water, the temperature rises to 50 degree Celsius. How much ice was in the container initially?
The heat capacity of water is 4.2 x 10^3
The heat capacity of ice is 2.1 x 10^3
The latent heat of vaporization is 2.3 x10^6

Homework Equations

Q =mc delta t
Q = mLv

The Attempt at a Solution

I didnt really have an attempt because i didnt know where to start, all i did was find the Q =mLv

 

[PeterO]

Homework Statement

An ice-water mixture has a mass of 180g and is contained in a 100g aluminum calorimeter When 35g of steam at 100 degree Celsius is condensed in the water, the temperature rises to 50 degree Celsius. How much ice was in the container initially?
The heat capacity of water is 4.2 x 10^3
The heat capacity of ice is 2.1 x 10^3
The latent heat of vaporization is 2.3 x10^6

Homework Equations

Q =mc delta t
Q = mLv

The Attempt at a Solution

I didnt really have an attempt because i didnt know where to start, all i did was find the Q =mLv

An ice/water mixture will be at 0 degrees.

What you have here is:

Some water heated up from 0 to 50,

Some ice melted, then heated up form 0 to 50

Some aluminium heated up from 0 to 50

Some steam condensed, then cooled from 100 to 50.

The first 3 required heat to happen,

The fourth gave up heat while happening.

The heat given up and the heat required are the same.

As for masses, you know the Al and the steam. The water and ice total 180.

 

[justinh8]

Ya I understand that part but how do i find the energy required for the Ice and Water if i don't have a mass? but only a mass for both of them together?

 

[PeterO]

Ya I understand that part but how do i find the energy required for the Ice and Water if i don't have a mass? but only a mass for both of them together?

Once the ice melts, you have 180g of water at 0 degrees.

100g of Al and 180g of water need just so much energy to reach 50 degrees.
the extra energy from the steam must have been used to melt the ice; so you can calculate how much ice there was.

 

[justinh8]

Oh, Ok i think i got it, i ended up with 138g of ice in the mixture initially which is correct according to the answers. Thanks!