Question

A copper block of mass 6.8 kg is originally at a temperature of 18°C and 1 atm. It is then heated to a temperature of 69°C without any change in pressure. The specific heat of copper is 387 J/(kg · °C), its density is 8.94 103 kg/m3, and the coefficient of linear expansion is 17 ✕ 10 −6 /°C.

(a) Determine the work done by the copper block. J

(b) How much heat energy is transferred during this process? J

(c) What is the increase in internal energy of the system? J

Answer 1

Part (a)

Given that copper block is heated at constant pressure, So work-done by copper block during isobaric process will be:

W = -P*dV

P = Pressure of copper block = 1 atm = 1.01325*10^5 N/m^2

dV = change in volume of block due to heating = V0**dT

V0 = Original Volume = Mass/density = 6.8 kg/(8940 kg/m^3) = 7.606*10^-4 m^3

= Volumetric expansion coefficient = 3

= linear expansion coefficient of copper = 17*10^-6

dT = change in temperature = 69 - 18 = 51 C

So,

W = -P*V0**dT

W = -1.01325*10^5*7.606*10^-4*3*17*10^-6*51

W = -0.20045 J (Work-done will be negative since we need work-done BY block)

Part (b)

Heat energy required during above process will be:

Q = m*C*dT

C = Specific heat capacity of copper = 387 J/kg.C

So,

Q = 6.8*387*51

Q = 134211.6 J = 1.342*10^5 J

Part C.

Increase in internal energy of the system will be given by, Using 1st law of thermodynamics:

dU = Q + W

dU = 134211.6 - 0.20045

dU = change in internal energy = 134211.39955 J = 134211.4 J

Let me know if you've any query.