please help me solve this.

Question

115,000 J of heat are added to a 82.7 g sample of ice @ -28.5 degrees celsius. Determine the final temperature and the grams of each phase present after the water absorbs all of the heat.

Answer 1

The first stage is to warm the ice to 0 °C.  The specific heat of ice is 2.11 J / g °C, so heating the ice to 0 would take 2.11 * 82.7 * 28.5 J = 4973.2 J.

 

The heat of fusion of water/ice is 334 J /g.  That's how much energy it takes to melt the ice per gram.  So that's 334 * 82.7 = 27621.8 J.

 

So warming and melting the ice has taken a total of 32595 J.  That leaves 82405 J.

 

The specific heat of water is 4.18 J / g °C, seating the water to 100 °C would take 4.18 * 82.7 * 100 J = 34568.6 J, so there is enough heat to do that.

 

That would leave 47836.4 J, which would be used to vaporize the water.

 

The heat of vaporization of water is 2260 J /g.  So there's enough heat left to vaporize 47836.4 / 2260 = 21.2 g, which would leave some water. The temperature would not rise about 100 until all the water is vaporized, so that means that the final temperature will be 100 °C.

 

There would be 21.2 g of steam and the rest, 61.5 g, would be water.