The complete combustion of 0.100 mol of CH4 in oxygen in a bomb calorimeter caused the water of the calorimeter to rise in temperature from 25.00 degrees Celsius to 37.70 degrees Celsius. Suppose that the combustion of the 0.1 mol of CH4 was instead run at constant pressure (1 atm) in a calorimeter that had a heat capacity (6973 J/degrees C) identical to that of the bomb calorimeter, and that also started at 25.00 degrees C; what would be the final temperature of the water in this calorimeter?
To find the final temperature of the water in the calorimeter, we can use the principle of conservation of energy. The heat released by the combustion of CH4 in both calorimeters will be the same. The two calorimeters have the same heat capacity, so they can absorb the same amount of heat.
The heat released by the combustion of CH4 can be calculated using the heat of combustion (ΔHc) of CH4. The balanced chemical equation for the combustion of CH4 is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that the stoichiometric ratio between CH4 and H2O is 1:2. This means that for every 1 mole of CH4 combusted, 2 moles of water are produced. Therefore, the heat released by the combustion of 0.100 mol of CH4 can be calculated as follows:
Q = ΔHc * n
Where:
Q = heat released (in joules)
ΔHc = heat of combustion of CH4 (in joules/mol)
n = number of moles of CH4 combusted
Assuming the heat of combustion of CH4 is -890 kJ/mol (this value can be found in thermodynamics tables), we can convert it to joules/mol:
ΔHc = -890 kJ/mol * 1000 J/kJ = -890,000 J/mol
Now we can calculate the heat released by the combustion of 0.100 mol of CH4:
Q = -890,000 J/mol * 0.100 mol = -89,000 J
Since the heat absorbed by the calorimeter is equal to the heat released by the combustion reaction, we can set up the following equation:
Q = mcΔT
Where:
Q = heat (in joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)
We need to find the change in temperature (ΔT) in this equation and use it to calculate the final temperature of the water.
Solving for ΔT:
ΔT = Q / (mc)
Substituting the known values into the equation:
ΔT = -89,000 J / (m * 4.18 J/g°C)
Since the mass of water is not given, we cannot calculate the exact final temperature without knowing it. However, we can still solve for the final temperature using a hypothetical mass. Let's assume a mass of 100 grams:
ΔT = -89,000 J / (100 g * 4.18 J/g°C) = -212.44°C
Since we started at 25.00°C, the final temperature would be:
Final temperature = 25.00°C + (-212.44°C) = -187.44°C
Note that this final temperature is not physically possible, as it is below the freezing point of water. To find the actual final temperature, we would need to know the mass of water in the calorimeter.