Question

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 8.60 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.

Answer 1

Answer: $64.216\times 10^4grams$ of liquid sodium is required.

Explanation: To calculate the mass of liquid sodium, we use the formula:

$q=nc\Delta T\\q=\frac{m}{M}c\Delta T$

where,

q = heat required, $8.60MJ=8.60\times 10^6J$   (Conversion factor: 1MJ = 1000000J)

m = Mass of liquid sodium, $?g$

M = Molar mass of liquid sodium, $23g/mol$

c = Specific heat capacity, $30.8J/Kmol$

$\Delta T$ = change in temperature,  $\Delta T=(510-500)K=10K$    (Conversion factor: 0°C = 273K)

Putting values in above equation, we get:

$8.60\times 10^6J=\frac{m}{23g/mol}\times 30.8J/Kmol\times 10K$

$m=64.216\times 10^4grams$