Mass ratio of steam-ice mixture

[Aaronkwg]

Homework Statement

Ice cube at 0 degree is mixed with steam at 100 degree, both become water at 50 degree.
Find the ratio of mass of the ice : mass of steam

Homework Equations

Q=ml
Q=mcdT

The Attempt at a Solution

I tried this way :
let the ratio of ice to steam = a:b
a x 3.34x10^5=b x 2.26x10^6
and i found out it was 6.8:1, but the answer is 4.5:1
I have no idea.
Thx in advance!

 

[eczeno]

you are ignoring the temperature changes. the ice melts to water at 0C and steam condenses to water at 100C, but then they come to equilibrium at 50C.

 

[Aaronkwg]

So should i add the 4200x50xa and 4200x50xb?
But I still get it wrong.

 

[eczeno]

basically you need to balance the energy; any energy that comes out one system goes into the other. so, on one side of the equation, you should record all the energy it takes to melt, and then raise the temperature of, the ice/water. on the other side, record the energy to condense, and then lower the temperature of, the steam/water. you seem to have all of the terms correct, and when i use your numbers i get 4.5:1 as well, so maybe check your arithmetic.

it would help if you put up all of your work: for instance 4200x50xa is the energy needed to raise the temperature of the water that was ice, but it is not clear how you used that to make your calculations.

 

[rwisz]

Thank you for providing your attempt at a solution. Your approach is on the right track, but it appears that you may have made a mistake in your calculation. Let's break down the problem and approach it step by step.

First, we need to determine the amount of heat required to melt the ice and vaporize the steam. This can be calculated using the specific heat capacity (c) and the latent heat of fusion (L) and vaporization (Lv) for water:

Qice = micexLice = (mice)x(334 J/g)
Qsteam = msteamxLvsteam = (msteam)x(2.26x10^6 J/kg)

Next, we can set up an equation using the heat gained by each substance and the final temperature (50 degrees Celsius):

Qice + Qsteam = micexcwaterx(50-0) + msteamxcwaterx(50-100)

Substituting in the values for Qice and Qsteam and solving for the ratio of mice to msteam, we get:

mice:msteam = (2.26x10^6 J/kg)/(334 J/g)x(50-100)/(50-0) = 4.5:1

Therefore, the correct ratio of mass of ice to mass of steam in the mixture is 4.5:1. It is possible that your mistake may have been in converting the units of specific heat capacity and latent heat of fusion/vaporization. Always be careful to check your units and make sure they are consistent throughout the calculation. I hope this helps!