exactly 11.2 ml of water at 23°c are added to a hot iron skillet. all of the water is converted into steam at 100°c. the mass of the pan is 1.5 kg and the molar heat capacity of iron is 25.19 j/(m·°c). what is the temperature change of the skillet
To determine the temperature change of the skillet, we need to calculate the amount of heat transferred between the water and the skillet. The equation for heat transfer is given by:
Q = m * c * ΔT
Where:
Q = amount of heat transferred (in joules)
m = mass of the substance (in kilograms)
c = specific heat capacity of the substance (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
First, let's calculate the amount of heat transferred from the water to the skillet. We'll use the equation:
Q_water = m_water * c_water * ΔT_water
Given:
m_water = 11.2 ml = 0.0112 kg (since 1 ml of water has a mass of approximately 1 g)
c_water = specific heat capacity of water = 4.18 J/(g·°C) = 4180 J/(kg·°C)
ΔT_water = 100°C - 23°C = 77°C
Substituting the values into the equation:
Q_water = 0.0112 kg * 4180 J/(kg·°C) * 77°C
Q_water = 3498.176 J (rounded to four decimal places)
Next, let's calculate the amount of heat absorbed by the skillet:
Q_skillet = m_skillet * c_skillet * ΔT_skillet
Given:
m_skillet = 1.5 kg
c_skillet = 25.19 J/(m·°C)
ΔT_skillet = ?
Since the water is being converted into steam at its boiling point of 100°C, we can assume that the skillet's temperature will also increase to 100°C. Therefore, the change in temperature for the skillet is:
ΔT_skillet = 100°C - Initial Temperature (let's assume the initial temperature is 23°C)
ΔT_skillet = 77°C
Substituting the values into the equation:
Q_skillet = 1.5 kg * 25.19 J/(m·°C) * 77°C
Q_skillet = 2928.885 J (rounded to four decimal places)
Finally, we need to determine the temperature change of the skillet. Since the water transferred heat to the skillet, we can equate the two equations:
Q_water = Q_skillet
3498.176 J = 2928.885 J
Now, let's solve for ΔT_skillet:
ΔT_skillet = Q_skillet / (m_skillet * c_skillet)
ΔT_skillet = 2928.885 J / (1.5 kg * 25.19 J/(m·°C))
ΔT_skillet = 77.413 °C
Therefore, the temperature change of the skillet is approximately 77.4°C.