Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 7.00 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than 10.0 °C? Use CP= 30.8 J/(K·mol) for Na(l) at 500 K.
Same type problem as hexane
To find the minimum amount of liquid sodium needed to absorb 7.00 MJ of energy while limiting the temperature increase to 10.0 °C, we can use the equation:
q = CP * M * ΔT
where:
- q is the heat absorbed by the sodium coolant
- CP is the molar heat capacity of sodium in J/(K·mol)
- M is the molar mass of sodium in g/mol
- ΔT is the desired temperature increase
First, we need to convert the given energy of 7.00 MJ to joules:
7.00 MJ = 7000000 J
Next, we need to calculate the amount of heat absorbed by the sodium, which is equal to the given energy:
q = 7000000 J
Now, let's substitute the values into the equation and solve for M:
q = CP * M * ΔT
7000000 J = (30.8 J/(K·mol)) * M * (10.0 °C)
We need to convert the temperature from Celsius to Kelvin by adding 273.15:
7000000 J = 30.8 J/(K·mol) * M * (10.0 + 273.15) K
Now, solve for M:
7000000 J = 30.8 J/(K·mol) * M * 283.15 K
Divide both sides by (30.8 J/(K·mol) * 283.15 K):
M = 7000000 J / (30.8 J/(K·mol) * 283.15 K)
M ≈ 825.18 mol
Lastly, convert moles to grams using the molar mass of sodium:
Molar mass of sodium (Na) ≈ 22.99 g/mol
Mass = M * molar mass
Mass = 825.18 mol * 22.99 g/mol
Mass ≈ 18,962.17 g
Therefore, a minimum of approximately 18,962.17 grams of liquid sodium is needed to absorb 7.00 MJ of energy while limiting the temperature increase to 10.0 °C.