Phase Changes Chemistry Problem? Can Someone Show me step by step how to do this.?

Question

The heat of fusion of water is 80 cal/g, the heat of vaporization of water is 540 cal/g, the specific heat of ice is 0.48 cal/deg/g, the specific heat of steam is 0.48 cal/deg/g and the specific heat of liquid water is 1.0 cal/deg/g. How much heat would be needed to convert 21.07 g of ice at -44 degrees C to steam at 141 degrees C.

Answer 1

On a lab bench in a non-pressurized (variable volume) vessel at standard atmospheric pressure:

21.07 * (44 *0.48 + 80 + 100 * 1 + 540 + 41 * 0.48) = 16030 cal
= about 16 Kcal

More precisely, since the isobaric specific heat of liquid water (cwp) and steam (cvp) are not constant over temperature, the real answer is calculated by integrating cwp and cvp over the temperature change. This more accurate value is 16083 cal

In a fixed volume container (isochoric), you will need to supply more or less heat depending on the volume. A 30 liter container will require the same amount of heat as the isobaric case. Volumes down to about 17 liters will require more heat, and volumes over 30 liters will require less.

Answer 2

Step 1. Find the heat for the temperature change in ice from -44 C to 0 C.

Step 2. Find the heat for the phase change from ice to water at 0 C.

Step 3. Find the heat for the temperature change from 0 C to 100 C. (water)

Step 4. Find the heat for the phase change from water to steam at 100 C

Step 5. Find the heat for the temperature change from 100 C to 141 C. (steam)

Step 6. Add Steps 1-5 together.

You should have the formulas to use for each of those 5 steps. If you need more help, let me know.

Answer 3

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