SOLUTION PHY101-6Th Assignment
QUESTION 1
A cube of cooper 2.00cm on a side is suspended by a string. The cube is
heated with a burner
from 20.0 C to 90.0 C. The air surrounding the cube is atmospheric pressure
(1.01 × 105Pa).
Find
a) The increase in volume of the cube;
b) The mechanical work done by the cube to expand against the pressure of
the surrounding air;
c) The amount of heat added to the cube;
d) The change in internal energy of the cube.
Note: For Copper = 5.1 × 10-3 (C )-1 Cp = 390J/Kg.K & (Rho) = 8.90 ×
103Kg/m3
SOLUTION:
a)
For small relative changes in volume and constant coefficient of thermal
expansion, this change
is given by:
V=V T
I think there is a typo in your question. The volumetric thermal expansion
coefficient for copper is
5.1×10 °C¹ and not 5.1×10³ °C¹.
Therefore
V = (2.0cm)³ 5.1×10 °C¹ (90°C - 20°C)
= 2.856×10² cm³
= 2.856×10 m³
b)
Work done by the copper cube to the surrounding is given by the integral
W = p dV form initial to final volume.
Since surrounding pressure is constant, the integral simplifies to
W = p dV = p
V
=>
W = 1:01×10 Pa 2.856×10 m³ = 2.88×10³ J
c)
Heat transferred to the cube in a constant pressure process equals the
change in enthalpy:
Q = H = m Cp T = V Cp T
=>
Q = 8.90×10³ kg/m³ (2.0×10² m)² 390 J/kg K 70K = 1943.76 J
d)
Change in internal energy equals transferred to the gas minus work done by
the gas:
U=Q-W
= 1943.76 J - 2.88×10³ J
= 1943.757 J
Actually the difference between U and H is small
2-ANSWER --SIMPLE AND STRAIGHT
As length=2cm=.02m
So volume =l*l*l=l^3
Vol =(.02)3=.0008m^3
T1=20C and T2=90C
T2-T1=90-20=70C
T2-T1=70+273=343K
Given that p=1.01*10^5 Pa
(a) increase in volume of cube=?
As V2 –V1= B(T2-T1)
V2-V1=5.1*10^-3 *.0008 *343
Increase in volume =133994.4*10^-7 m3
(b) work done=?
W=P*deltaV
W=P(V2-V1)
W=1.01*10^5 *133994.4 * 10^-7
W=14134.344 *10^-2
W=141.34J
(c) heat added=?
At constant pressure,Cp=Qp/deltaT
Qp=Cp *delta T
Qp=390 *343
=133770 J
(d) as Q=delta U +W
deltaU=Q-W
=133770-141.343
=133628.637