Question

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 9.50 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than $10.0^{o}C$ ? use CP = 30.8 J/K.mol.

Answer 1

There is a formula having a relation between heat, mass and change in temperature of the chemical reaction and it is as follows.
$Q=mc({{T}_{2}}-{{T}_{1}})$
Here Q = the amount of energy absorbed
m = mas of the metal
c = heat capacity
${{T}_{2}}-{{T}_{1}}$ = change in temperature.

Complete answer:
- In the question it is given that to calculate how many grams of liquid sodium needed to absorb 9.50 MJ of energy.
- The relation between heat capacity, amount of energy absorbed, temperature change and mass of the metal is as follows.
$Q=mc({{T}_{2}}-{{T}_{1}})$
Here Q = the amount of energy absorbed $= 2.30\times {{10}^{6}}$
m = mas of the metal
c = heat capacity = $\dfrac{{{C}_{M}}}{M}=\dfrac{{{C}_{M}}}{23}$
${{T}_{2}}-{{T}_{1}}$ = change in temperature $= 10^{o}C$
- Substitute all the values in the above formula to get the amount of the required metal and it is as follows.

$$Q=mc({{T}_{2}}-{{T}_{1}}) \\\ \Rightarrow m=\dfrac{Q}{c({{T}_{2}}-{{T}_{1}})} \\\ \Rightarrow m=\dfrac{2.30\times {{10}^{6}}\times 23}{30.8\times 10} \\\ \Rightarrow m=1.72\times {{10}^{5}}=172kg \\\$$

- Therefore the mass of the liquid sodium is 172 kg.

Note: We need heat capacity, amount of energy absorbed, and temperature change to calculate the mass of the metal otherwise we cannot solve the problem. We should know the relationship between them also to get the exact value of the liquid metal required to cool the machine engine.