You are a blacksmith and have been working with 14.0 kg of steel. When you are finished shaping it, the steel is at a temperature of 405°C. To cool it off, you drop it into a bucket containing 3.0 kg of water at 55°C. How much of this water is converted to steam? Assume the steel, the water, and the steam all have the same final temperature.
To determine the amount of water that is converted to steam, we need to calculate the heat transferred from the steel to the water and then use that to find how much water vaporizes.
The heat transferred from the steel to the water can be calculated using the formula:
Q = m * c * ΔT
Where,
Q = Heat transferred (in Joules)
m = mass of the substance (in kg)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
For steel, the specific heat capacity (c) is approximately 450 J/kg°C, and the mass (m) is 14.0 kg. The initial temperature (Ti) of the steel is 405°C, and the final temperature (Tf) is the same as the water's temperature (55°C). Therefore the change in temperature (ΔT) for the steel is:
ΔT = Tf - Ti = 55°C - 405°C = -350°C
Now we can calculate the heat transferred from the steel to the water:
Q = 14.0 kg * 450 J/kg°C * (-350°C)
Q = -2,205,000 J
The negative sign indicates that heat is transferred from the steel to the water.
The heat required to convert a certain mass of water into steam at its boiling point is given by the formula:
Q = m * Lv
Where,
Q = Heat transferred (in Joules)
m = mass of water converted to steam (in kg)
Lv = latent heat of vaporization (in J/kg)
The latent heat of vaporization for water is approximately 2,260,000 J/kg.
Now we can calculate the amount of water converted to steam:
-2,205,000 J = m * 2,260,000 J/kg
m = -2,205,000 J / 2,260,000 J/kg
m ≈ -0.975 kg
Since mass cannot be negative, we take the absolute value:
m ≈ 0.975 kg
Therefore, approximately 0.975 kg of water is converted to steam.