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I'm not very sure of when should be used the heat of fusion and the heat of condensation for water. This happened when I attempted to solve a problem of this matter.

The problem is as follows:

A flask whose heat capacity is equal to $1.8\,\frac{cal}{^{\circ}C}$ has $89\,g$ of ice inside, all of it at $0^{\circ}C$. Then $30\,g$ of steam is injected at $100^{\circ}C$ 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:

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

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 $80\frac{cal}{g}$

Using the formula of

$\textrm{heat gained = - heat lost}$

$q_{ice}+q_{melted}=-q_{steam}$

$80\times 89+89\times 1 \times (x-0)= 30\times 1 \times (100-x)$

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

$x=-34.622$

which doesn't make sense as the temperature can't fall that low. Can someone guide me with what should I do here?. I'm confused whether should I had used the heat of condensation as the steam will condense, but what will be this value?. $540\frac{cal}{g}$? but positive or negative?. Can someone help to clear out this?. I'm still unable to solve this situation. Help please!

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The heat balance equation should read:

Heat gained by ice and flask = Heat lost by steam

Heat gained by ice and flask = $(89)(80)+(89)(1)(T-0)+(1.8)(T-0)$

Heat lost by steam = (30)(540)+(30)(1)(100-T)

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  • $\begingroup$ The mass of the flask is unknown, and phase changes are involved (not shown in your equation for heat gained by ice and flask). This means that the given question can't be answered. $\endgroup$ – David White Feb 23 at 18:47
  • $\begingroup$ @DavidWhite The mass times the specific heat capacity of the flask is specified in the problem statement as1.8 cal/C, and accounted for in the expression for the heat gained by ice and flask.. The phase change in going from ice to liquid water at 0 C is accounted by by the first term in the expression for the heat gained by Ice and flask. The phase change for the condensation of the steam is accounted for in the first term in the expression for the heat lost by the steam. $\endgroup$ – Chet Miller Feb 23 at 19:04

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