500 calories of heat is added to 2.0 g of water at 80 degrees Celsius. How much steam is produced?
2.0 * 20 = 40 calories will be needed to raise the temperature to the boiling point of 100 C. That will leave 460 calories for vaporization. The heat of vaporization of water is 540 cal/g.
460/540 g will vaporize.
To find out how much steam is produced when 500 calories of heat is added to 2.0 g of water at 80 degrees Celsius, we need to use the concept of specific heat and latent heat of vaporization.
First, let's calculate the amount of heat needed to raise the temperature of 2.0 g of water from 80 degrees Celsius to its boiling point. The specific heat capacity of water is 1 calorie/gram °C.
q1 = m * c * ΔT
q1 = 2.0 g * 1 calorie/gram °C * (100 °C - 80 °C)
q1 = 2.0 g * 1 calorie/gram °C * 20 °C
q1 = 40 calories
So, it takes 40 calories of heat to raise the temperature of 2.0 g of water from 80 degrees Celsius to its boiling point.
Next, we need to calculate the amount of heat required to vaporize the water. The latent heat of vaporization for water is 540 calories/gram.
q2 = m * L
q2 = 2.0 g * 540 calories/gram
q2 = 1080 calories
So, it takes 1080 calories of heat to vaporize 2.0 g of water.
Now, let's add up the heat required for temperature increase and vaporization:
Total heat = q1 + q2
Total heat = 40 calories + 1080 calories
Total heat = 1120 calories
Since 500 calories of heat is added, we can find out how much steam is produced by dividing the added heat by the total heat:
Mass of steam = (500 calories / 1120 calories) * 2.0 g
Mass of steam = 0.446 g
Therefore, approximately 0.446 grams of steam is produced when 500 calories of heat is added to 2.0 g of water at 80 degrees Celsius.