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.
To determine the minimum amount of liquid sodium needed to absorb 7.00 MJ of energy, we need to use the equation:
Q = m × C × ΔT,
where:
Q is the amount of energy absorbed (in joules),
m is the mass of the liquid sodium (in grams),
C is the specific heat capacity of the liquid sodium (in J/(g·K)),
ΔT is the change in temperature (in degrees Celsius).
First, we need to convert the energy from MJ to J:
7.00 MJ = 7.00 × 10^6 J.
Next, we rearrange the formula to solve for the mass of liquid sodium:
m = Q / (C × ΔT).
Now, we can substitute in the known values:
Q = 7.00 × 10^6 J,
C = 30.8 J/(K·mol) (specific heat capacity of sodium),
ΔT = 10.0 °C (maximum change in temperature).
However, notice that the specific heat capacity given is in J/(K·mol), so we need to perform an additional conversion to get the specific heat capacity in J/(g·K) since we want the mass in grams.
To convert from molar specific heat capacity to specific heat capacity:
CP (J/(K·mol)) × Molar mass (g/mol) = C (J/(g·K)).
The molar mass of sodium is approximately 22.99 g/mol.
C = 30.8 J/(K·mol) × 22.99 g/mol = 708.47 J/(g·K).
Now, we can substitute this value into our formula for mass:
m = Q / (C × ΔT) = (7.00 × 10^6 J) / (708.47 J/(g·K) × 10.0 °C).
Now, let's calculate the mass:
m = (7.00 × 10^6 J) / (708.47 J/(g·K) × 10.0 °C).
m = (7.00 × 10^6 J) / (7084.7 g·K).
m ≈ 987.3 grams.
Therefore, a minimum of approximately 987 grams of liquid sodium is needed to absorb 7.00 MJ of energy, ensuring the temperature does not increase by more than 10.0 °C.