Question

An iron block of 12 kg undergoes a process during which there is a heat gain from the block at 2 kJ/kg, an elevation increase of 32 m, and a decrease in velocity from 40 m/s to 7 m/s. During the process, which also involves work transfer, the internal energy of the block increases by 70 kJ. Suppose the total energy of the system remains constant. Determine the work transfer during the process in kJ and indicate whether the work is done on/by the system.

Answer 1

Step-by-step explanation:

Total heat gain by the block ΔQ = 2 x 12 kJ = 24 kJ .

Gain of potential energy = mgh = 12 x 9.8 x 32 = 3.763 kJ

Decrease in kinetic energy KE = 1/2 x 12 ( 40² - 7² )

= 9.306 kJ

increase in internal energy ΔE = 70 kJ

ΔQ = ΔE + PE - KE + W , W is work done by the gas

Putting the values

24 = 70 + 3.763 - 9.306 + W

W = - 40.457 kJ .

Since W is negative that means work is done on the system .