91Ó°ÊÓ

The system undergoes a process with energy transfer through heat at a constant rate of 10 kW. The power \( \dot{W} \) varies with time as follows: \[ \dot{W} = \begin{cases}\-8t & 0 < t \leq 1 \mathrm{h} \ \ \ \ \ \ -8 & t > 1 \ \mathrm{h}\end{cases} \] Given \( \dot{W} \) is in kW and \( t \) is in hours.

The rate of energy transfer by heat is \(Q = 10 \mathrm{kW} \). At \( t = 0.6 \mathrm{h} \), the power \( \dot{W} = -8t \). For \( t = 0.6 \mathrm{h} \), \ \( \dot{W} = -8 \cdot 0.6 = -4.8 \mathrm{kW} \). The time rate of change of system energy is given by \( \frac{dE_{\rm system}}{dt} = \dot{Q} - \dot{W} \). Substituting the values we get: \ \( \frac{dE_{\rm system}}{dt} = 10 - (-4.8) = 10 + 4.8 = 14.8 \mathrm{kW} \).

The total change in system energy can be determined by integrating the rate of change over the time periods. For 0 < t \leq 1 \mathrm{h}, \ \(\ \frac{dE_{\rm system}}{dt} = 10 - (-8t) = 10 + 8t\ . \) Integrate \(10 + 8t \) from 0 to 1 hour: \[ \Delta E_{\rm system,1} = \int_0^1 (10 + 8t) \ dt = \[ 10t + 4t^2 \]_{0}^{1} = 10 \cdot 1 + 4 \cdot 1^2 = 10 + 4 = 14 \ \mathrm{kWh} \]. For t > 1 hour: \ \(\ \frac{dE_{\rm system}}{dt} = 10 - (-8) = 10 + 8 = 18 \ \mathrm{kW} \) \ (constant). Integrate 18 from 1 to 2 hours: \[ \Delta E_{\rm system,2} = \int_1^2 18 \ dt = 18t \[ 1 \to 2 \] = 18 \cdot 1 = 18 \ \mathrm{kWh} \]. Total change in system energy after 2 hours: \ \(\ \Delta E_{\rm system,tot} = 14 \ \mathrm{kWh} + 18 \ \mathrm{kWh} = 32 \[ \mathrm{kWh} \] = 32 \cdot 3600\ \ve 115200 \mathrm{kJ} \).