Return to Blackboard cnnhns Moran, Shapiro, Boettner, Baliley, Fundamentals of E

Question

Return to Blackboard cnnhns Moran, Shapiro, Boettner, Baliley, Fundamentals of Engineering Thermodynamics, Se Help I System An Periodi 4 days left I Register Ne Assignment Gradebook Downloadable eTextbook gnment BACK Probiem 7.044 A system of mass 12 kg undergoes a prodess dunng which there is no work, the elevation decreases ty 50 m, ans the velooty increases from 15 m's to 60 m/s. The specfic internal energy decreases by 5 kjkg and the acceleration of gravity is constant at 9.8 ms Determine the change in kinetic energy, in kd, and the amount of energy transfer by heat for the process, in ka. Step 1 Determine the change in kinetic energy, in k KE 12 the tolerance is +/2% Click if you would like to Show Work for this questions en. 5hon won sewr re"LATER Attempts: 0 of 15 used Step 2 The parts of this question must be completed in ordier. Tha part wil te available when yow complete the part abowe ,, 4 3

Explanation / Answer

Kinetic energy = 1/2m*(v^2)

Velocity changes from 15m/s to 60m/sec

mass of the system = 12kg

So,change in kinetic energy due to change in speed = 1/2*12*(60^2) - 1/2*12*(15^2)

=21600-1350

=20250 Joules

=20.25 KJ

Since there is no work done on the system,

Total energy transfered due to heat+ total change in energy in descending 50 meters =0.

Change in potential energy = mass*g*(change in height)  

=12*9.8*(-50)

= -5880 J

= -5.88KJ

Change in internal energy = -5KJ/kg

Total change in internal energy = -5*12 = -60KJ

Total change in energy = (change in Kinetic energy+change in Potential Energy+change in internal energy)

=(20.25- 5.88- 60)KJ

= - 45.63KJ

Hence total heat tranfered to the system = -(-45.63)= 45.63 KJ

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