- #1

TheStebes

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**[SOLVED] Latent heat, steam and ice question**

## Homework Statement

14g of steam at 100C is added to 47g of ice at 0.0 degrees C.

a) Find the amount of ice melted and the final temperature.

## Homework Equations

Q=mL

Q=mc[tex]\Delta[/tex]T

Q(hot)=-Q(cold)

## The Attempt at a Solution

Ordinarily, in such a problem, you would have to consider the various processes which affect Q(hot) and Q(cold). However, in this problem, the starting temperatures are 100 and 0 C, so the heat or energy goes directly to the phase change, correct?

m[tex]_{steam}[/tex]=.014kg

m[tex]_{ice}[/tex]=.047kg

L[tex]_{vaporization}[/tex]=2.26*10^6

L[tex]_{fusion}[/tex]=3.33*10^5

c[tex]_{water}[/tex]=4186

Q(hot)= -m[tex]_{s}[/tex]*L[tex]_{v}[/tex] + m[tex]_{s}[/tex]*c[tex]_{w}[/tex]*T[tex]_{f}[/tex]

Q(cold)= m[tex]_{i}[/tex]*L[tex]_{f}[/tex] + m[tex]_{i}[/tex]*c[tex]_{w}[/tex]*T[tex]_{f}[/tex]

set

Q(hot)=-Q(cold)

solving for T[tex]_{f}[/tex], I get 62.62 degrees C.

Obviously I'm misusing or neglecting one or more equations in the Q calculations, as my solution does not even solve for the mass of the water. I'm thoroughly confused as to which equation to use with which variables. (between latent heat and the specific heat capacity equation.)