591_Physics ProblemsTechnical Physics

# 591_Physics ProblemsTechnical Physics - Chapter 20 P20.57...

This preview shows page 1. Sign up to view the full content.

Chapter 20 593 P20.57 Qm cT V cT == ∆∆ ρ bg so that when a constant temperature difference T is maintained, the rate of adding energy to the liquid is P F H G I K J = dQ dt dV dt R ρρ and the specific heat of the liquid is c RT = . P20.58 (a) Work done by the gas is the negative of the area under the PV curve WP V V PV i i i ii =− F H G I K J =+ 22 . (b) In this case the area under the curve is d V z . Since the process is isothermal, PV P V nRT i i i F H G I K J = 4 4 and W dV V V V V V i i i i F H G I K J F H G I K J = z 4 4 4 ln ln 139 . FIG. P20.58 (c) The area under the curve is 0 and W = 0. P20.59 Call the initial pressure
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online