Question

A heat recovery system​ (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loops use Freon as the working fluid. The instrumentation must be kept as a temperature greater than negative 67−67 degrees Fahrenheit​ [°F] to avoid damage. The temperature in the area of Mars where the rover is exploring is 189189 kelvins​ [K]. If the system must remove 47.447.4 British Thermal Units​ [BTU] of​ energy, what volume of Freon is needed in units of liters​ [L]?

Answer 1

`V=2.0757* 10^(-4) * 1000 = 0.20757 \,liter`

Step-by-step explanation:

Given that:

quantity of heat to be removed,
`Q= 47.4\,BTU`

we know,

`1\,BTU= 1055.06\,J`

density of freon,
`\rho=1490\,kg.m^(-3)`

latent heat of vaporization of freon,
`L=161.7* 10^3 J.kg^(-1)`

So,

heat to be removed in joules,

`Q=47.4* 1055.06`

`Q=50009.844\,J`

Now, the quantity of freon required to remove the above mentioned heat:

`m=(Q)/(L)`

`m=(50009.844)/(161.7* 10^3)`

`m=0.3093\,kg`

Now, volume

`V=(m)/(\rho)`

`V=(0.3093)/(1490)`

`V=2.0757* 10^(-4)m^3`

`V=2.0757* 10^(-4) * 1000`

`V = 0.20757 \,liter`