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 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
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`
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`
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`
it is c.........