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A vertical piston-cylinder device contains water and is being heated. During the heating process, 70 kJ of heat is transferred to the water, and heat losses from the side walls amount to 8 kJ. The piston rises as a result of evaporation, changing the volume by 0.5 m3. When the atmospheric pressure is 1 atm and the mass of the piston is 50 kg and its cross-sectional area is 0.05 m2, determine the change in the energy of the water for this process.

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Answer:

Energy change = 59. 5 kJ

Step-by-step explanation:

First, we collect the data:

Let:

Q (supplied) be the energy supplied to the water

Q (lost) be energy lost to surroundings

Q (piston) be the energy possessed by the piston as it moves up

v = volume flow (m³/s)]

V = velocity of the gas (m/s)

mass = 50 kg

Second, calculate the work done by the piston:

to find velocity, we use this formula:

v = A × V

Therefore, V = 0.5 (m³/s)/0.005 m²

= 10 m/s

it means the water is evaporating at a rate of 10 m/s

the evaporating water transfers energy to the piston, kinetic energy Ek

Ek = 1/2 mV²

= 0.5 (50) ( 10) ²

= 2 500 J

= 2.5 kJ

The net energy, ΔQ is given by:

ΔQ (net) = Q (supplied) - Q (piston) - Q (lost)

= 70 - 2.5 -8

= 59.5 kJ Ans

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