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A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned 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 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.

Respuesta :

Answer:

Explanation:

Heat absorbed by oil

= mass x specific heat x rise in temperature

= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )

= 25.08 x 10⁵ J  

Heat absorbed by air

= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )

= 6.03 x 10⁵ J

Total heat absorbed = 31.11 x 10⁵ J

If time required = t

heat lost from room

= .35 x 10³ t

Total heat generated in time t

= 1.8 x 10³ t

Heat generated = heat used

1.8 x 10³ t =  .35 x 10³ t  + 31.11 x 10⁵

1.45 x 10³ t = 31.11 x 10⁵

t = 31.11 x 10⁵ / 1.45 x 10³

t = 2145.5 s

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