Liquid sodium is being considered as an engine coolant. how many grams of liquid sodium (minimum) are needed to absorb 9.50 mj 9.50 mj of energy in the form of heat if the temperature of the sodium is not to increase by more than 10.0 °c? 10.0 °c? use c p = 30.8 j / ( k ⋅ mol ) cp=30.8 j/(k⋅mol) for na ( l ) na(l) at 500 k .

709415.584 g

Explanation:

This is a heat capacity  problem. We use the formula below to calculate.

Q=mcΔT

where Q is the quantity of heat

           m is the mass of substance

           ΔT is the change in temperature of substance( T2-T1).

Q= 9.5MJ = 9.5*10^(6}J

C.P = 30.8J/(K.mol)

temperature of sodium increases by 10°C which means our final temperature in kelvin T2 will be 510K

initial temperature T1= 500K

m= Q/[c (T2-T1)]

m= 9.5*10^(6}/ {30.8(510-500)]

m= 30844.156 mol of Na

Converting to grams we multiply by the molar mass of Na.

molar mass of Na = 23

mass of Na=   30844.156 *23

                  =709415.584 g

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