Physics, 10.03.2020 04:48 adrianbanuelos1999

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 65 degrees Fahrenheit [°F] to avoid damage. The temperature in the area of Mars where the rover is exploring is 192 kelvins [K]. If the system must remove 46.9 British Thermal Units [BTU] of energy, what volume of Freon is needed in units of liters [L]?

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Answer from: GreenHerbz206

Volume of Freon = 12.23. $10^{3}$ L

Explanation: First, the temperatures have to be in the same unit, so all the tmeperatures is transformed into Kelvin:

T₁ = 192K

T₂ = $\frac{5}{9}$ . (F - 32)+273.15

T₂ = $\frac{5}{9}$ . (- 65 - 32)+273.15

T₂ = 219.26K.

Second, BTU is an unit of energy. It can be transformed into Joules by the relation 1BTU = 1055.06J.

Q = 1055.06 . 46.9 = 49482.314J

The letter Q represents Heat and is calculated as Q = m.Cp.ΔT, where:

m is mass of the element;

Cp is the heat capacity specific for each element;

ΔT is the difference between the final and initial temperature;

Third, it will be needed the properties of the Freon:

Freon (CF2Cl2) = 120.91g/mol; Cp = 74J/mol.K; ρ = 1.49kg/m³

Fourth, we calculate the mass of Freon necessary for the remove of 46.9BTU of energy from the system:

Q = m.Cp.ΔT

m= $\frac{Q}{c(T-T_{0} )}$

m = $\frac{49482.314}{74(219.26-192)}$

m = 18228.214g or m=18.228Kg

Fifth, using the density of Freon, Volume can be found

ρ = $\frac{m}{V}$

V = m / ρ

V = 18.228 / 1.49

V = 12.23m³

As SI for density is Kg/m³, the volume found is in m³.

Cubic meters is related to Liters as the following: 1m³=1000l.

So, volume of Freon = 12.23. $10^{3}$ L

The volume of Freon needed is 12.23. $10^{3}$ L.

Answer from: Quest

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2 b

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explanation:

Answer from: Quest

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A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mar...