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A 40-L electrical radiator containing heating oil is
Chapter 4, Problem 38P(choose chapter or problem)
A \(40-\mathrm{L}\) electrical radiator containing heating oil is placed in a \(50-\mathrm{m}^{3}\) room. Both the room and the oil in the radiator are initially at \(10^{\circ} \mathrm{C}\). The radiator with a rating of \(2.4 \mathrm{~kW}\) is now turned on. At the same time, heat is lost from the room at an average rate of \(0.35 \mathrm{~kJ} / \mathrm{s}\). After some time, the average temperature is measured to be \(20^{\circ} \mathrm{C}\) for the air in the room, and \(50^{\circ} \mathrm{C}\) for the oil in the radiator. Taking the density and the specific heat of the oil to be \(950 \mathrm{~kg} / \mathrm{m}^{3}\) and \(2.2 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, determine how long the heater is kept on. Assume the room is well-sealed so that there are no air leaks.
Equation Transcription:
10°C
20°C
50°C
°C
Text Transcription:
40-L
50-m^3
10 degree calsius
2.4 kW
0.35 kJ/s
20 degree calsius
50 degree calsius
950 kg/m^3
2.2 kJ/kg degree calsius
Questions & Answers
QUESTION:
A \(40-\mathrm{L}\) electrical radiator containing heating oil is placed in a \(50-\mathrm{m}^{3}\) room. Both the room and the oil in the radiator are initially at \(10^{\circ} \mathrm{C}\). The radiator with a rating of \(2.4 \mathrm{~kW}\) is now turned on. At the same time, heat is lost from the room at an average rate of \(0.35 \mathrm{~kJ} / \mathrm{s}\). After some time, the average temperature is measured to be \(20^{\circ} \mathrm{C}\) for the air in the room, and \(50^{\circ} \mathrm{C}\) for the oil in the radiator. Taking the density and the specific heat of the oil to be \(950 \mathrm{~kg} / \mathrm{m}^{3}\) and \(2.2 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, determine how long the heater is kept on. Assume the room is well-sealed so that there are no air leaks.
Equation Transcription:
10°C
20°C
50°C
°C
Text Transcription:
40-L
50-m^3
10 degree calsius
2.4 kW
0.35 kJ/s
20 degree calsius
50 degree calsius
950 kg/m^3
2.2 kJ/kg degree calsius
ANSWER:
Step by step solution. Step 1 of 3In this problem we need to calculate how long the heater was kept on.To do that westart by determining the total heat the heater needed to generate .The heater was heating up the radiator and the room .To determine the heatneeded for the radiator . We will need the change in temperature of the heater,specific heat of the oil and the mass of the oil .The temperature change we can calculate from the initial and final temperature of the radiator.\nThe mass of the oil we can calculate by using the given volume of the radiator and the density of oil .Before the calculation we need toexpress the volume in units.We can now calculate the heat needed for the radiator