A 135 g - g sample of steam at 100 ∘C is emitted from a volcano. It condenses, cools, and falls as snow at 0 ∘C . (For water, 80. cal (334 J ) is needed to melt 1 g of ice or must be removed to freeze 1 g of water and 540 cal (2260 J ) is needed to convert 1 g of water to vapor at 100 ∘C .)
How many kJ of heat were released?
AAAaannndd the bot gets it wrong yet again!
2.26 kJ per gram. You do not have 1 gram.
@oobleck your unreal man
To find out how many kJ of heat were released, we need to determine the amount of heat released during the phase changes of the steam.
First, let's calculate the heat released when the steam condenses and cools to water at 100 °C.
The heat released during condensation is given by the formula:
Q = mass × latent heat of vaporization
The mass of the steam is given as 135 g, and the latent heat of vaporization of water at 100 °C is 540 cal/g (2260 J/g). We need to convert the result to kJ, so we'll divide by 1000.
Q1 = (135 g) × (540 cal/g) ÷ 1000 = 72.9 kJ
Now, to find the heat released when the water further cools and freezes into ice at 0 °C, we'll use the formula:
Q = mass × latent heat of fusion
The mass of the water is still 135 g, and the latent heat of fusion of water is given as 80 cal/g (334 J/g). Again, we'll divide the result by 1000 to get kJ.
Q2 = (135 g) × (80 cal/g) ÷ 1000 = 10.8 kJ
Finally, to find the total heat released, we'll add the two heat values together:
Total heat released = Q1 + Q2
= 72.9 kJ + 10.8 kJ
= 83.7 kJ
Therefore, approximately 83.7 kJ of heat were released during the phase changes of the steam.