If 5.40 kcal of heat is added to 1.00 kg of water at 100⁰C, how much steam at 100⁰C is produced? Show all calculations leading to an answer.
Is this how you would do this?
Q= m*Hvap
Q= m* 5.40
Q=100*5.40
Steam = 540 kg of steam?
You need to use units.
Hv=2260kj/kgwater=2260kj*1kcal/4.18kj*1/kgwater
Hv= 540kcal/kgwater.
Q=m*Hv
mass=Q/Hv= 5.40kg/(540kcal/kgwater)
= .01kg=100grams water
0.01kg = 10.0 grams water.
To calculate how much steam at 100°C is produced when 5.40 kcal of heat is added to 1.00 kg of water at 100°C, you need to use the equation Q = m * Hvap.
Q represents the heat added to the water, m represents the mass of water, and Hvap represents the heat of vaporization.
In this case, Q = 5.40 kcal and m = 1.00 kg.
To solve for Hvap, you'll need to use the specific heat capacity of water, which is 1 kcal/kg°C, and the boiling point of water, which is 100°C.
The equation for Hvap is:
Hvap = m * Cp * ΔT
Where Cp represents the specific heat capacity and ΔT represents the change in temperature. Since the water is at its boiling point, ΔT = 100°C - 0°C = 100°C.
Therefore, Hvap = 1.00 kg * 1 kcal/kg°C * 100°C = 100 kcal
Now that we have the value for Hvap, we can substitute it back into the original equation and solve for m, which represents the mass of steam produced.
Q = m * Hvap
5.40 kcal = m * 100 kcal
Dividing both sides of the equation by 100 kcal:
m = 5.40 kcal / 100 kcal
Thus, the mass of steam produced is:
m = 0.054 kg
Therefore, when 5.40 kcal of heat is added to 1.00 kg of water at 100°C, approximately 0.054 kg of steam at 100°C is produced.
So, the correct calculation is:
Steam = 0.054 kg of steam.