A small vertical panel heater for a room loses heat by convection and radiation combined, according to the following equation: Qout = (he + ha). A. AT where Qaut is the total heat flowing out from the panel in watts, AT is the temperature difference between the panel surface and the surroundings in "C, and A is the surface area of the panel in m?. The heat transfer coefficient h in W/(m? "C) is split into two parts: he for the convective part of the heat loss, and ha for the radiative part of the heat loss. For small temperature differences, the radiative heat transfer coefficient hr can be assumed to be fairly constant at 5 W/(m?, "C). However, for the convective part, the heat transfer coefficient he is itself a mild function of temperature because the heated air against the panel rises in convective currents that flow faster at a higher temperature. To a first approximation, it can be assumed that he = 2. (AT)025 The panel of the heater has a mass of 5 kg, and it can be assumed that its average specific heat is 800 J/kg."C. The surface area of the panel is 1.0 m', and it can be assumed that its entire surface is at a uniform temperature although this temperature will vary with time. Write a dynamic heat balance for the heater, and linearise this dynamic heat balance around the steady-state operating point when the power into the heater is 400 W, and rearrange this into the normal LDE form with the terms in the output variable on the left side.