Re: (VU-Study-Corner) Discussion PHY101 Assignment !....

Thanks to share assignment

On Tue, Jul 12, 2011 at 2:31 PM, mc100401703 Fouzia Tabassum <mc100401703@vu.edu.pk> wrote:

SOLUTION PHY101-6

Th 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

On Tue, Jul 12, 2011 at 11:38 AM, Ahsan Siddiqui <bc090403083@gmail.com> wrote:

please do send last assignment solution file to me

Course: PHY101_physics

thanks in advance

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