Exactly 10 mL of water at 25°C was added to a hot iron skillet. All of the water was converted into steam at 100°C. If the mass of the pan was 1.20 kg and the molar heat capacity of iron is 25.19 J/mol∙°C, what was the temperature change of the skillet?
To determine the temperature change of the skillet, we can use the principle of heat transfer. The heat gained by the skillet is equal to the heat lost by the water.
First, let's calculate the heat gained by the skillet using the equation:
Q = m * c * ΔT
Where:
Q is the heat gained by the skillet,
m is the mass of the skillet,
c is the molar heat capacity of iron, and
ΔT is the temperature change of the skillet.
Substituting the given values:
Q = (1.20 kg) * (25.19 J/mol∙°C) * ΔT
Now, let's calculate the heat lost by the water. Since all the water is converted into steam, we can use the equation:
Q = m * ΔH_vap
Where:
Q is the heat lost by the water,
m is the mass of the water, and
ΔH_vap is the heat of vaporization for water.
The heat of vaporization for water is 40.7 kJ/mol or 40.7 J/g.
Since we know the volume of water is 10 mL and the density of water is 1 g/mL, the mass of water can be calculated as:
m = V * ρ
Where:
m is the mass of water,
V is the volume of water, and
ρ is the density of water.
Substituting the given values:
m = (10 mL) * (1 g/mL) = 10 g
Now, let's calculate the heat lost by the water:
Q = (10 g) * (40.7 J/g) = 407 J
Since the heat gained by the skillet is equal to the heat lost by the water, we can equate the two equations:
(1.20 kg) * (25.19 J/mol∙°C) * ΔT = 407 J
Simplifying the equation:
25.19 ΔT = 407 / (1.20 kg)
ΔT = (407 / (1.20 kg)) / 25.19
Calculating ΔT:
ΔT = 13.2 °C
Therefore, the temperature change of the skillet is 13.2 degrees Celsius.