Question

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per kg, an elevation decrease of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific internal energy decreases by 5 kJ/kg and the acceleration of gravity is constant at 9.7 m/s2. Determine the heat transfer for the process, in kJ.

Answer 1

Answer:-50.005 kJ

Explanation:

Given

mass of system =10 kg

work done=0.147 kJ/kg

Change in elevation$(\Delta h)=-50 m$

initial velocity $(v_1)=15 m/s$

Final Velocity$(v_2)=30 m/s$

Specific internal Energy$(\Delta U)=-5 kJ/kg$

from first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE= kinetic energy

PE=potential energy

U=internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q=1.47+3.375-4.850-50

Q=-50.005 kJ