CHANGE OF STATE PRACTICE
ANSWER SHEET
First we have to remember that there are two changes of state, and each one has its own latent heat. First the change of state between a solid and a liquid has a latent heat of 80 calories per gram. Second the change of state between a liquid and a gas has a latent heat of 600 calories per gram. Next we should remember that the calories needed for changing the temperature, also called the specific heat, is 1 calorie per gram – O C. The temperatures for each change of state are: solid to liquid occurs at 0 O C, while the change from liquid to gas occurs at 100 O C.
If you look at the units of the answer, you will see what you have to do to get there!!!!
1. How many calories does it take to raise the temperature of 1 gram of water by 10O C?
Here we are changing temperature only. Therefore we only use the specific heat.
Therefore Calories = calories/gram - O C = (1 cal/gm/ O C) (10 O C) (1 gm) = 10 calories
2. How many calories are needed to convert 5 grams of ice to water?
Now we have a change in state, from solid to liquid, without any change in temperature.
Therefore: Calories = (80 cal/gm) (5 gm) = 400 calories
3. How many calories are needed to convert 3 grams of water to water vapor?
Once again we have a change in state, from liquid to gas.
Therefore: Calories = (600 cal/gm) ( 3 gm) = 1800 calories
4. How many calories are needed for 1 gram of H2O to go from -10 O C to + 110 O C?
Here we have two changes in state, from solid to liquid and from liquid to gas. We also have a change in temperature. You may solve this problem the long way by taking each step at a time. Starting with the temperature change, then the change in state from solid to liquid, followed by a temperature change and a change in state from liquid to gas, and completed by the last temperature change.
OR
We can combine both changes of state, and all of the temperature changes. It does not matter which one we use as the math works out the same for each. When we combine the temperature changes remember that - degrees are added to the + degrees to get the total.
LONG WAY
Temperature change from - 10 to 0
(1 cal/gm/OC) (10O C) (1 gm) = 10 cal
Change in state from solid to liquid
(80 cal/gm) (1 gm) = 80 cal
Temperature change from 0 to 100
(1 cal/gm/ OC) ( 100O C) (1 gm) = 100 cal
Change in state from liquid to gas
(600 cal/gm) (1 gm) = 600 cal
Temperature change from 100 to 110
(1 cal/.gm/O C) ( 10O C) ( 1 gm) = 10 cal
total = 800 calOR THE EASY WAY
Temperature change from -10 to +110 or 120 degrees
(1 cal/gm/O C) (120O C) (1 gm) = 120 cal
Change of state from solid to gas
(680 cal/gm) ( 1 gm) = 680 cal
total = 800 cal
5. How many calories are involved in changing 5 grams of H2O from + 125 O C to – 2 O C?
This is a similar problem in reverse. Since the latent heats are the same no mater which way you go, we use the same values.
EASY WAY
Temperature change from + 125 to -2 degrees or 127 degrees
(1 cal/gm/O C) (127O C) ( 5 gm) = 635 cal
Change in state from gas to solid
(680 cal/gm) ( 5 gm) = 3400 cal
total = 4035 calLONG WAY
Temperature change from +125 to +100 degrees
(1 cal/gm/ OC) ( 25O C) (5 gm) = 125 cal
Change of state from gas to liquid
(600 cal/gm) ( 5 gm) = 3000 cal
Temperature change from +100 to 0 degrees
(1 cal/gm/ OC) ( 100O C) (5 gm) = 500 cal
Change of state from liquid to solid
(80 cal/gm) (5 gm) = 400 cal
Temperature change from 0 to -2 degrees
(1 cal/gm/ OC) ( 2O C) (5 gm) = 10 cal
answer = 4035 cal