Question

A system of mass 10 kg undergoes a process during which there is no work, the elevation decreases by 50 m, and the velocity increases from 15 m/s to 50 m/s. The specific internal energy decreases by 5 kJ/kg and the acceleration of gravity is constant at 9.8 m/s2. Determine the change in kinetic energy, in kJ, and the amount of energy transfer by heat for the process, in kJ.

Answer 1

Answer:

Change in kinetic energy: 11375 J

Heat transfer energy: 43525 J

Explanation:

The change in kinetic energy can be calculated using the following formula:

$\Delta E_K = mv_2^2/2 - mv_1^2/2 = \frac{m}{2}(v_2^2 - v_1^2)$

$\Delta E_K = 10/2(50^2 - 15^2) = 11375 J$

The change in internal energy is

$\Delta E_i = -5 * 10 = -50 kJ = -50000 J$

The change in potential energy is

$\Delta E_p = mgh_2 - mgh_1 = mg(h_2 - h_1) = mg\Delta h = 10*98*(-50) = -4900$

Let $E_h$ be the heat transfer in the process, according to law of energy conservation, all of this energy change must be balanced out. Therefore:

$\Delta E_k + \Delta E_i + \Delta E_p + E_h = 0$

$11375 - 50000 - 4900 + E_h = 0$

$E_h = 50000 + 4900 - 11375 = 43525 J$