A 40-L electrical radiator containing heating oil is placed in a 50-m³ room. Both the room and the oil in the radiator are initially at 10°C. The radiator with a rating of 2.4 kW is now turned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.

After some time, the average temperature is measured to be 20°C for the air in the room, and 60°C for the oil in the radiator.

Taking the density and the specific heat of the oil to be 950 kg/m³ and 2.2 kJ/kg-°C, respectively, determine how long the heater is kept on.

Assume the room is well sealed so that there are no air leaks. The gas constant of air is [tex]R = 0.287 \text{ kPa-m}^3/\text{kg-K}[/tex] (Table A-1). Also, [tex]c = 0.718 \text{ kJ/kg-K}[/tex] for air at room temperature (Table A-2). Oil properties are given to be [tex]p = 950 \text{ kg/m}^3[/tex] and [tex]C_p = 2.2 \text{ kJ/kg-°C}[/tex].

The heater is kept on for ______ minutes.

To determine how long the heater is kept on, we need to calculate the heat gained by the room and the heat lost from the room. The heat gained by the room can be calculated using the equation Q = mcΔT, where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The heat lost from the room can be calculated using the equation Q = PΔt, where Q is the heat lost, P is the power, and Δt is the time. Equating both equations, we can solve for Δt.

The heat gained by the room is given by Q = mcΔT = (density of air) x (specific heat of air) x (volume of room) x ΔT

The heat lost from the room is given by Q = PΔt = (average rate of heat loss) x Δt

Setting both equations equal to each other and solving for Δt, we find that the heater should be kept on for approximately 47,338 seconds, or about 12.4 hours.