91Ó°ÊÓ

The First Law for a closed system can be written as \[\frac{dU}{dt} = \dot{Q} + \dot{W}\] where \(\frac{dU}{dt}\) is the rate of change of internal energy, \(\dot{Q}\) is the rate of heat transfer, and \(\dot{W}\) is the rate of work done by the paddle wheel.

For an incompressible liquid with constant specific heat, the change in internal energy can be expressed as \[\Delta U = mc\Delta T\]where \(m\) is the mass, \(c\) is the specific heat, and \(T\) is the temperature. Then, the rate of change of internal energy can be expressed as \[\frac{dU}{dt} = mc\frac{dT}{dt}\]

Given the heat transfer rate \[\dot{Q} = -hA(T - T_0)\] and assuming the paddle wheel adds energy at a constant rate \(\dot{W}=W_0\), substitute these into the First Law equation: \[mc\frac{dT}{dt} = W_0 - hA(T - T_0)\]

Reorganize the terms to form: \[mc\frac{dT}{dt} + hA(T - T_0) = W_0\] Define the parameters \(\tau = \frac{mc}{hA}\) and \(C = \frac{W_0}{hA}\): \[\frac{dT}{dt} + \frac{1}{\tau}(T - T_0) = \frac{C}{\tau}\]

This is a first-order linear differential equation. Solve it using the integrating factor method. The integrating factor is \[\mu(t) = e^{\int \frac{1}{\tau} dt} = e^{\frac{t}{\tau}}\] Multiply the differential equation by \(\mu(t)\): \[e^{\frac{t}{\tau}}\frac{dT}{dt} + \frac{1}{\tau}e^{\frac{t}{\tau}}T = \frac{C}{\tau}e^{\frac{t}{\tau}}\] This simplifies to: \[\frac{d}{dt}(e^{\frac{t}{\tau}} T) = \frac{C}{\tau} e^{\frac{t}{\tau}}\] Integrate both sides with respect to \(t\): \[e^{\frac{t}{\tau}} T = C e^{\frac{t}{\tau}} + D\] where \(D\) is a constant of integration. Solve for \(T\): \[T = C + De^{-\frac{t}{\tau}}\] Given initial condition at \(t=0\), \(T(0) = T_0\): \[T_0 = C + D\] Thus, \[D = T_0 - C\]. Substitute back to find \(T(t)\): \[T(t) = C + (T_0 - C)e^{-\frac{t}{\tau}}\]

Since \(C = \frac{W_0}{hA}\), the final form of the temperature equation is: \[T(t) = \left(\frac{W_0}{hA}\right) + \left(T_0 - \frac{W_0}{hA}\right)e^{-\frac{t}{\tau}}\]