Boiling of water in an open container is a constant pressure process. On earth water boils...

Boiling of water in an open container is a constant pressure process. On earth water boils at 100 deg C (atmospheric pressure is 101.3 kPa). If we were to boil 2 kg of water on the surface of Mars, how much energy (Q) do we need to supply? The ΔU for this process is 4750 kJ. Specific volumes of water vapor and liquid are 206.00 and 0.00100 m3/kg, respectively.

Expert Solution

Q= dH-W

Q= ?

Mass of water : 2Kg

Specific volume of water vapour Vg= 206

Specific volume of water Vf = 0.001

Vg-Vf = 206-0.001 = 206 approximately

Since water is boiling in an open container therefore W=0

Q=dH

dH = dU + d(PV)

ΔU= 4750 kJ

dH= 4750 + (101.325 × 206) = 25622.95 KJ/Kg

total energy require Q to boil the water = 25622.95 × mass of the water on the mars

= 25622.95× 2Kg

= 51245.9 KJ/Kg

queen_honey_blossom answered 2 years ago

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