Answer:
The final temperature of the air is [tex]T_2= 605 K[/tex]
Explanation:
We can start by doing an energy balance for the closed system
[tex]\Delta KE+\Delta PE+ \Delta U = Q - W[/tex]
where
[tex]\Delta KE[/tex] = the change in kinetic energy.
[tex]\Delta PE[/tex] = the change in potential energy.
[tex]\Delta U[/tex] = the total internal energy change in a system.
Q = the heat transferred to the system.
W = the work done by the system.
We know that there are no changes in kinetic or potential energy, so [tex]\Delta KE = 0[/tex] and [tex]\Delta PE=0[/tex]
and our energy balance equation is [tex]\Delta U = Q - W[/tex]
We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.
We use this information to calculate the total internal energy change [tex]\Delta U=W+Q[/tex] using the energy balance equation.
We convert the interval of time to seconds [tex]t = 5 \:min = 300\:s[/tex]
[tex]\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t[/tex]
[tex]\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ[/tex]
We can use the change in specific internal energy [tex]\Delta U = m(u_2-u_1)[/tex] to find the final temperature of the air.
We are given that [tex]T_1=300 \:K[/tex] and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy [tex]u_1[/tex]
[tex]u_1=214.07\:\frac{kJ}{kg}[/tex]
Next, we will calculate the final specific internal energy [tex]u_2[/tex]
[tex]\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1[/tex]
[tex]\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}[/tex]
[tex]u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}[/tex]
With the value [tex]u_2=439.07 \:\frac{kJ}{kg}[/tex] and the ideal gas tables for air we make a regression between the values [tex]u = 434.78 \:\frac{kJ}{kg},T=600 \:K[/tex] and [tex]u = 442.42 \:\frac{kJ}{kg}, T=610 \:K[/tex] and we find that the final temperature [tex]T_2[/tex] is 605 K.
