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?
The temperature change of the skillet can be calculated using the equation for the heat capacity of a substance, which is $q=mc\Delta T$, where $q$ is the heat, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.
In this case, the heat absorbed by the skillet is the heat of vaporization of water, which is 40.7 kJ/mol. Since the skillet has a mass of 1.20 kg and the molar heat capacity of iron is 25.19 J/mol∙°C, the temperature change can be calculated as follows:
$\Delta T = \frac{q}{mc} = \frac{40.7 \text{ kJ/mol}}{1.20 \text{ kg} \cdot 25.19 \text{ J/mol∙°C}} = 1.61 \text{ °C}$
Therefore, the temperature of the skillet increased by 1.61°C when the water was added.
Well, turning water into steam usually requires some heat, but turning an iron skillet into a steak is a whole different story! But let's focus on the question, shall we?
To find the temperature change of the skillet, we need to calculate the heat absorbed by the iron skillet. To do that, we can use the equation:
q = m × c × ΔT
Where:
q is the heat absorbed,
m is the mass of the iron skillet,
c is the molar heat capacity of iron, and
ΔT is the temperature change.
Since we know the mass of the skillet is 1.20 kg, the molar heat capacity of iron is 25.19 J/mol∙°C, and we want to find ΔT, we can rearrange the equation:
ΔT = q / (m × c)
Now, let's find q. The heat absorbed by the skillet can be calculated by multiplying the amount of water converted into steam by the heat required to vaporize water, which is the molar heat of vaporization of water:
q = n × ΔHvap
We know that the molar heat capacity of an object is the product of its molar mass and its specific heat capacity, so:
c = M × Cp
Rearranging the equation to solve for M:
M = c / Cp
Now, we know that 10 mL of water is converted into steam, and the molar mass of water is approximately 18 g/mol. So, we can calculate the moles of water (n) as follows:
n = volume / molar mass
In this case, volume = 10 mL = 0.01 L.
n = 0.01 L / 0.018 kg/mol
Now, we can calculate ΔHvap using the formula:
q = n × ΔHvap
So, q = n × ΔHvap = m × c × ΔT
Finally, plugging in the known values into the equation:
ΔT = q / (m × c)
Sorry, I got a bit carried away there. It seems like the skillet might have gotten a bit hotter than expected! Let's calculate the temperature change, shall we?
To calculate the temperature change of the skillet, we need to use the heat equation:
Q = m × c × ΔT
Where:
Q is the heat transferred
m is the mass of the object (in this case, the skillet)
c is the specific heat capacity of the object (in this case, the iron)
ΔT is the change in temperature
We know that all of the water was converted into steam, so the heat transferred can be calculated using the mass of the water and the heat of vaporization of water.
The heat required to convert water to steam is given by:
Q = m × Hv
Where:
Q is the heat transferred
m is the mass of the water
Hv is the heat of vaporization of water
Considering the values:
m of water = 10 mL = 10 grams (1 ml ≈ 1 gram for water)
Hv of water = 40.7 kJ/mol = 40,700 J/mol
Now we can calculate the heat transferred to convert the water to steam:
Q = m × Hv
Q = 10 g × (40,700 J/mol)
Q = 407,000 J
Next, we can calculate the heat transferred to the iron skillet:
Q = m × c × ΔT
407,000 J = 1.20 kg × 25.19 J/mol∙°C × ΔT
To find ΔT, we rearrange the equation:
ΔT = Q / (m × c)
ΔT = 407,000 J / (1.20 kg × 25.19 J/mol∙°C)
Calculating the value:
ΔT = 407,000 J / 30.228 J/°C
ΔT ≈ 13.47 °C
Therefore, the temperature change of the skillet is approximately 13.47 °C.
To find the temperature change of the skillet, we need to calculate the heat absorbed by the skillet. We can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed by the skillet,
m is the mass of the skillet,
c is the molar heat capacity of iron, and
ΔT is the temperature change.
First, let's calculate the heat absorbed by the water as it converts into steam. The heat absorbed by the water can be calculated using the equation:
Q = m × ΔHvap
Where:
m is the mass of the water, and
ΔHvap is the heat of vaporization of water, which is 40.7 kJ/mol.
Given that we added exactly 10 mL of water, we need to convert this to grams:
10 mL × 1 g/mL = 10 g
Now that we know the mass of the water, we can calculate the heat absorbed by the water:
Q = (10 g) × (40.7 kJ/1000 g) = 0.407 kJ
Since 1 kJ = 1000 J, the heat absorbed by the water is:
Q = 0.407 kJ × 1000 J/1 kJ = 407 J
Now, let's calculate the heat absorbed by the skillet:
Q = mcΔT
407 J = (1200 g) × (25.19 J/mol·°C) × ΔT
Rearranging the equation to solve for ΔT:
ΔT = 407 J / [(1200 g) × (25.19 J/mol·°C)]
ΔT ≈ 0.014 °C
Therefore, the temperature change of the skillet is approximately 0.014 °C.