Please i need help in this question thank you.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...

Please i need help in this question thank you.

A cube of cooper 2.00cm on a side is suspended by a string. The cube is heated with a burner from 20.0C to 90.0C. 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 β = 5.1 × 10-3 (C)-1 Cp = 390J/Kg.K & ρ (Rho) = 8.90 × 103Kg/m3

Asked on by sonog4

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valentin68 | College Teacher | (Level 3) Associate Educator

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The volume of the cube is

`V = d^3 =0.02^3 =8*10^-6 m^3`

The coefficient of volumetric expansion for copper is `gamma=5.1*10^-6 K^-1`

The increase in temperature is `Delta(T) =90-20 =70 K =70 @C`

The law of volume variation with temperature is

`Delta(V) =gamma*V*Delta(T) =5.1*10^-6*8*10^-6*70 =2.856*10^-9 m^3`

`Delta(V) =2.856*10^-3 cm^3`

b)

The mechanical work done by the cube to increase its volume against the  atmospheric pressure is

`W =P*Delta(V) =1.01*10^5*2.856*10^-9 =2.885*10^-4 J`

c)

The mass of the cube is

`m = rho*V = 8900*8*10^-6 =7.12*10^-2 kg =71.2 grams`

The heat necessary to increase the temperature of the cube is

`Q =m*Cp*Delta(T) =7.12*10^-2*390*70=1943.76 J`

d) If `U` is the internal energy of the cube then the first principle of thermodynamics says:

`Q =Delta(U) +W`

`Delta(U) =Q-W =1943.76-2.885*10^-4 =1943.7597 J`

Sources:

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