Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of heat is transferred to the water and 5 kJ of heat is lost to the surrounding air. The paddle-wheel work amounts to 500 N·m. Determine the final energy of the system if its initial energy is 19.5 kJ.

U₂=45 kJ

Explanation:

Given Data

Initial Energy U₁=19.5 kJ

Qin=30 kJ

Win=500 N. m

Qout = 5 kJ

To find

Final energy U₂

Solution

From The first law of thermodynamic

ΔE = Ein - Eout + ΔQ - ΔW

or

U₂=U₁+Qin+Win-Qout

where

U₂ is final energy

U₁ is initial energy

q = energy transferred as heat to a system

w = work done on a system

U₂ = (19.5+30+500*10⁻³-5) kJ

U₂=45 kJ