Question

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.20 MJ 1.20 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 .

Answer 1

Answer : The mass of liquid sodium needed are, $3.17\times 10^3g$

Explanation :

First we have to calculate the number of moles of liquid sodium.

$q=n\times c\times (\Delta T)$

where,

q = heat absorb = 1.20 MJ = 1.20 × 10⁶ J

n = number of moles of liquid sodium = ?

c = specific heat capacity = $30.8J/mol.K$

$\Delta T$ = change in temperature = $10.0^oC=273+10.0=283K$

Now put all the given values in the above formula, we get:

$1.20\times 10^6J=n\times 30.8J/mol.K\times 283K$

$n=137.7mol$

Now we have to calculate the mass of liquid sodium.

Molar mass of Na = 23 g/mol

$\text{Mass of Na}=\text{Moles of Na}\times \text{Molar mass of Na}$

$\text{Mass of Na}=137.7mol\times 23g/mol=3167.1g=3.17\times 10^3g$

Therefore, the mass of liquid sodium needed are, $3.17\times 10^3g$