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A brass cube, 10 cm on a side, is heated with a temperature change of 200 deg. C. By what percentage does its volume change?
Vo = 10cm = 1000cm^3 = .001m^3
T = 200 C
brass = 19x10^-6 coefficient of linear expansion
V=Vo(1+3 x brass x T)
or
change in V = Vo(3 x brass x T)
= .001m^3(3 x 19x10^-6 x 200 C)
= .0000114 = 1.14x10^-5
% change = change in V/Vo
= .0000114/.001
= .0114 = 1.1%
answer in the book is 1.1x10^-3 %
What I am doing wrong here?
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A radial tire is inflated to a gauge pressure of 35 lb/in^2 at 60 deg. F.
If the temperature increases to 100 deg F while the volume of the
tire remains constant, what is the tire's new pressure?
T1=60 F = 289 K
T2=100 F = 311 K
V1=V2
P1= 35 lb/in^2
Find P2
P1/T1=P2/T2
P2=P1T2/T1
P2 = (35 lb/in^2 x 311 K)/289 K
P2 = 37.7 lb/in^2
Also tried this:
P=pa + pg
14.7 (atm) + 35 lb/in^2
= 49.7 lb/in^2 = absolute pressure
P2 = (49.7 lb/in^2 x 311 K)/289 K
P2 = 53.5 lb/in^2
answer in the book is 39 lb/in^2
any help would be appreciated
Vo = 10cm = 1000cm^3 = .001m^3
T = 200 C
brass = 19x10^-6 coefficient of linear expansion
V=Vo(1+3 x brass x T)
or
change in V = Vo(3 x brass x T)
= .001m^3(3 x 19x10^-6 x 200 C)
= .0000114 = 1.14x10^-5
% change = change in V/Vo
= .0000114/.001
= .0114 = 1.1%
answer in the book is 1.1x10^-3 %
What I am doing wrong here?
-----------------------------------------------------------------------------------
A radial tire is inflated to a gauge pressure of 35 lb/in^2 at 60 deg. F.
If the temperature increases to 100 deg F while the volume of the
tire remains constant, what is the tire's new pressure?
T1=60 F = 289 K
T2=100 F = 311 K
V1=V2
P1= 35 lb/in^2
Find P2
P1/T1=P2/T2
P2=P1T2/T1
P2 = (35 lb/in^2 x 311 K)/289 K
P2 = 37.7 lb/in^2
Also tried this:
P=pa + pg
14.7 (atm) + 35 lb/in^2
= 49.7 lb/in^2 = absolute pressure
P2 = (49.7 lb/in^2 x 311 K)/289 K
P2 = 53.5 lb/in^2
answer in the book is 39 lb/in^2
any help would be appreciated