[Blueberries116]

The problem is as follows:

A flask whose heat capacity is equal to [itex]1.8\,\frac{cal}{^{\circ}C}[/itex] has [itex]89\,g[/itex] of ice inside, all of it at [itex]0^{\circ}C[/itex]. Then [itex]30\,g[/itex] of steam is injected at [itex]100^{\circ}C[/itex] in the flask. Find the amount of liquid water in the flask when the thermal equilibrium is reached. Assume the loss of energy to the environment is negligible.

The alternatives given are as follows:

[itex]\begin{array}{ll}
1.&\textrm{90 g}\\
2.&\textrm{96 g}\\
3.&\textrm{99 g}\\
4.&\textrm{119 g}\\
5.&\textrm{107 g}\\
\end{array}[/itex]

I'm confused exactly how should I approach this question. Initially I thought to do a heat balance as follows:

For purposes of brevity I'm omitting units but they're accordingly to their respective quantities. I'm working with grams and calories and using the latent heat of fusion of water to be [itex]80\frac{cal}{g}[/itex]

Using the formula of

[itex]\textrm{heat gained = - heat lost}[/itex]

[itex]q_{ice}+q_{melted}=-q_{steam}[/itex]

[itex]80\times 89+89\times 1 \times (x-0)= 30\times 1 \times (100-x)[/itex]

From solving this equation I obtained the temperature in the equilibrium to be:

[itex]x=-34.622[/itex]

which doesn't make sense as the temperature can't fall that low. Can someone guide me with what should I do here?.  Help please!

[Borek]

Without checking calculation details (which may, or may not, contain additional problems) - what about a heat of condensation?

[Enthalpy]

Additional problems, yes.

Why assume that all ice has melted?
Why assume that all vapour has condensed?
Condensed water is a liquid too, it contributes to the "amount of liquid water".

Here there are several states, I don't expect one equation to represent all possibilities. One must first determine what states will be present at equilibrium. Only after, a temperature can be determined.