^{6}J / 1 MJ ) } / { ( 30.8 J / mol•K ) x ( 10.0 K ) }

^{5}grams of Na

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 8.30 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.

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David L. | Chemistry TutorChemistry Tutor

Great question, for a universe with no oxidizing agents strong enough to react with the sodium. (Note: that ain't our universe.) Anywayz...

"q = mCΔT" is always a good place to start when looking at absorption of heat from a substance in a single phase.

q = the heat absorbed or released from the substance

m = the mass of the substance

C = the specific heat of the substance

ΔT = the change in temperature of the substance

Rearranging to find mass of liquid sodium, we can plug in the numbers, make sure the units cancel out to leave only grams, then calculate. To get the units right, it helps to check what units are used in the specific heat, i.e., Joules, Kelvin, and moles. (Sodium is 22.99 g/mol, a 10.0ºC temperature change is the same as a 10.0 K temperature change, and a megaJoule is a million Joules.)

m = q / {C x ΔT}

m = { 8.30 MJ x ( 10^{6} J / 1 MJ ) } / { ( 30.8 J / mol•K ) x ( 10.0 K ) }

x (22.99 g Na / 1 mol Na)

= 6.20 x 10^{5} grams of Na

Stanton D. | Tutor to Pique Your Sciences InterestTutor to Pique Your Sciences Interest

I would use the unit-factor method to solve this question, but formulas are easier to write about. The relevant formula to use is:

Q = (CP)(delta-T)(n) where Q is the heat energy transferred into or out of a material, CP is the heat capacity of the material, delta-T is the temperature change, and n is the number of moles. If the heat capacity were given on a mass-basis (as a specific heat capacity of a specific heat) then the mass m would be used instead of the number of moles n.

In this case, we're told that the temperature of sodium can't increase by more than 10.0 °C. A change of 10.0 °C is the same as a change of 10 K. We need to convert megajoules to Joules. Solving for n yields:

n = Q/((CP)(delta-T)) = (8.30X10^{6} J)/((30.8 J/(K·mol))(10.0 K)) = 2.69X10^{4} moles of Na .

Convert to grams of Na for your final answer.

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