Asystem consisting of 3 lb of water vapor in a piston–cylinder assembly, initially at 350°f and a volume of 71.7 ft3, is expanded in a constant‐pressure process to a volume of 85.38 ft3. the system then is compressed isothermally to a final volume of 28.2 ft3. during the isothermal compression, energy transfer by work into the system is 72 btu. kinetic and potential energy effects are negligible. determine the heat transfer, in btu, for each process.

isobaric expansion = 281.09 Btu

isothermal compression= 72 Btu

Explanation:

The first law of thermodynamics is:

where:

Q=heat transferred

W= work

U=internal energy  

P=pressure, V= volume, T= temperature, n =  moles, Cv= specific heat at constant volume.

In a isobaric process heat transferred is:

For an isothermal process (T2-T1 = 0) so

From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.

For the first process

we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)

we use this information in the appropriate unit to find the pressure in thermodynamic tables.

T1= 176°C

v1= 1.49 m^3/kg

P=1.37 bar

in the second state we have

P=1.37 bar =137000Pa

with thee properties we check in the thermodynamic tables

T2= 255°C

we usually find Cp on tables for water but from the Mayer relation we have:

Cp for water vapor is: 33.12 J/mol*K

R=8.314 J/mol*K

Cv= 41.434 J/mol*K

replacing in the equation for Q

296569J =281.09 Btu

answer:

a

explanation:

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