Question

A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. What is the change in internal energy of this control mass?

(Consider the tank and the fluid inside a control surface)

1. -3590 kJ
2. +3590 kJ
3. +4590 kJ
4. -4590 kJ
5.

Answer 1

Correct Answer - Option 2 : +3590 kJ

Concept:

The First Law of Thermodynamics for a process (state 1 to state 2) is given by-

Q1-2 = ΔU + W1-2

Q1-2 = (U2 - U1) + W1-2

Calculation:

Given:

W_input = 5090 kJ (negative),

Q_out = 1500 kJ (negative)

First Law of Thermodynamics:

Q1-2 = (U2 - U1) + W1-2

-1500 = (U2 - U1) + [-5090]

(U2 - U1) = -1500 + 5090 ⇒ +3590 kJ