1-27 A room is heated by an electrical resistance heater placed in a short duct in the room in 15 min while theroom is losing heat to the outside, and

a 300-W fan circulates the air steadily through the heater duct. The powerrating of the electric heater and the temperature rise of air in the duct are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical pointvalues of -141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, Ake = Ape =0. 3Constant specific heats at room temperature can be used for air. This assumption results in negligible error inheating and air-conditioning applications. 3 Heat loss from the duct is negligible. 4 The house is air-tight and thusno air is leaking in or out of the room. Properties The gas constant of air is R = 0.287 kPa.m³/kg.K (Table A-1). Also, c,= 1.007 kJ/kg-K for air at roomtemperature (Table A-15) and cu=cp-R = 0.720 kJ/kg-K. Analysis (a) We first take the air in the room as the system. This is a constant volume closed system since no masscrosses the system boundary. The energy balance for the room can be expressed as \underline{E_{\mathrm{b}}-E_{\mathrm{out}}}=\underbrace{\Delta E_{\mathrm{ggtr}}} W_{\mathrm{ein}}+W_{\mathrm{fin}}-Q_{\mathrm{du}}=\Delta U \left(\dot{W}_{\mathrm{Gin}}+\dot{W}_{\mathrm{fin}, \mathrm{in}}-\dot{Q}_{\mathrm{out}}\right) \Delta t=m\left(u_{2}-u_{1}\right) \cong m c_{v}\left(T_{2}-T_{1}\right) y=5 \times 6 \times 8 \mathrm{~m}^{3}=240 \mathrm{~m}^{3} m=\frac{P_{1} U}{R T_{1}}=\frac{(98 \mathrm{kPa})\left(240 \mathrm{~m}^{3}\right)}{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K}\right)(288 \mathrm{~K})}=284.6 \mathrm{~kg}