Exercise $3.-$ In the previous exercise, now we relax the assumption that energy is transferred instantaneously

from the heating element to the contents of the tank. Suppose that the metal heating element has a significant

thermal capacitance $C_e$ and that the electrical heating rate $Q(t)$ directly affects the temperature of the element

rather than the liquid contents. For simplicity, we neglect the temperature gradients in the heating element that

result from heat conduction and assume that this element has a uniform average temperature $T_e(t).$ Based on

these and the previous assumptions, the energy balances for the tank and the heating element can be written as

$mC\frac{dT}{dt}=wC(T_i-T)+h_e{A}_e(T_e-T)$

$m_e{C}_e\frac{d{T}_e}{dt}=Q-h_e{A}_e(T_e-T)$

where $m=V,$ and $m_e{C}_e$ is the product of the mass of metal in the heating element and its specific heat. The

term $h_e{A}_e$ is the product of the heat transfer coefficient and area available for heat transfer. $Q(t)$ is an input

variable, the thermal equivalent of the instantaneous electrical power dissipation in the heating element.

$($a$)$ Determine the steady$-$state equilibrium relationships among the different variables.

$($b$)$ Calculate the corresponding linearized equations around these steady$-$state values.

$($c$)$ Assuming that the flow rate $w(t)$ is constant, convert the previous equations into a single second$-$order

differential equation that involves only variables $T(t),T_i(t),Q(t)$ and $($possibly$)$ their derivatives.

Hint: Solve the first equation for $T_e$ and then differentiate.