Question

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m^3. The tank is fitted with a paddle wheel that transfers energy to the CO at a constant rate of 14 W for 1 h. During the process, the specific internal energy of the carbon monoxide increases by 10 kJ/kg. If no overall changes in kinetic and potential energy occur,
determine:
a. the specific volume at the final state;
b. the energy transfer by work;
c. the energy transfer by heat transfer, and the direction of the heat transfer.

Answer 1

Answer:

(a). The specific volume at the final state is 0.25 m³/kg.

(b). The energy transfer by work is 50.4 kJ.

(c). The energy transfer by heat  is -10.4 kJ.

Negative sign shows the direction of heat

Explanation:

Given that,

Weight of carbon monoxide = 4 kg

Volume of tank = 1 m³

Constant rate of 14W for 1 hour

The specific internal energy of the carbon monoxide increases by 10 kJ/kg.

(a). We need to calculate the specific volume at the final state

Using formula of specific volume

$V'=\dfrac{V}{m}$

Put the value into the formula

$V'=\dfrac{1}{4}$

$V'=0.25\ m^3/kg$

(b). We need to calculate the energy transfer by work

Using formula of work

$W=Pt$

Where, P = power

t = time

Put the value into the formula

$W=14\times1\times3600$

$W=50.4\ kJ$

(c). The energy transfer by heat transfer, and the direction of the heat transfer

We need to calculate the $\Delta U$

Using formula of internal energy

$\Delta U=m\Delta u$

Put the value into the formula

$\Delta U=4\times10$

$\Delta U=40\ kJ$

We need to calculate the energy transfer by heat

Using formula of energy transfer

$Q=\Delta U-W$

Put the value into the formula

$Q=40-50.4$

$Q=-10.4\ kJ$

Negative sign shows the direction of heat that is removed from CO.

Hence, (a). The specific volume at the final state is 0.25 m³/kg.

(b). The energy transfer by work is 50.4 kJ.

(c). The energy transfer by heat  is -10.4 kJ.

Negative sign shows the direction of heat.