How many calories are released when 3 grams of 100º C steam turns to 0º C ice
first turn steam to water
m * heat of vaporization
second turn water at 100 to water at 0
m * specific heat of water * 100
third turn water at 0C to ice
m * heat of fusion
add those
To determine the number of calories released when 3 grams of 100ºC steam turns to 0ºC ice, we need to calculate the heat lost during this phase change.
First, we need to find the heat energy required to cool the steam from 100ºC to 0ºC. The specific heat capacity of water is approximately 1 calorie/gram · ºC. Therefore, the heat energy required to cool 3 grams of steam from 100ºC to 0ºC is:
Q1 = mass × specific heat capacity × temperature change
= 3g × 1 cal/g·ºC × (100ºC - 0ºC)
= 300 calories
Next, we need to determine the heat energy released during the phase change of the steam turning into ice at 0ºC. This amount is known as the heat of fusion, which is the heat energy required per gram to change a substance from a liquid to a solid phase at its melting point. For water, the heat of fusion is approximately 79.7 calories/gram.
The heat energy released during the phase transition from steam to ice is given by:
Q2 = mass × heat of fusion
= 3g × 79.7 cal/g
= 239.1 calories
Therefore, the total calories released when 3 grams of 100ºC steam turns to 0ºC ice is:
Total heat released = Q1 + Q2
= 300 calories + 239.1 calories
= 539.1 calories
So, when 3 grams of 100ºC steam turns into 0ºC ice, approximately 539.1 calories are released.