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.

Short answer

Answer: The heater was kept on for approximately 32 minutes.

Step-by-step solution

Determine the Energy Supplied by the Radiator to the Room

To find the energy supplied by the radiator, we need to multiply the power rating of the radiator by the time for which it is on. Let's denote the time as \(t\). The power rating of the radiator is given as \(2.4\,\mathrm{kW}\), which is equal to \(2400\,\mathrm{J/s}\). Therefore, the energy supplied by the radiator is \(E_\text{supplied}= 2400t\,\mathrm{J}\).

Determine the Energy Gained by the Oil in the Radiator

To find the energy gained by the oil inside the radiator, we'll use the equation \(E_\text{oil}=mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature. The initial temperature of the oil is given as \(10\,^{\circ}\mathrm{C}\) and the final temperature is \(50\,^{\circ}\mathrm{C}\). Therefore, \(\Delta T = 50 - 10 = 40\,^{\circ}\mathrm{C}\). The mass of oil in the radiator can be found using the equation \(m=\rho V\), where \(\rho\) is the density of oil and \(V\) is the volume of oil in the radiator. The density \(\rho\) is given as \(950\,\mathrm{kg/m^3}\) and the volume \(V\) as \(40\,\mathrm{L}\), which equals \(0.04\,\mathrm{m^3}\). Hence, mass \(m = 950 \times 0.04 = 38\,\mathrm{kg}\). The specific heat capacity of the oil, \(c\), is given as \(2.2\,\mathrm{kJ/kg\cdot^{\circ}\mathrm{C}}\), which equals \(2200\,\mathrm{J/kg\cdot^{\circ}\mathrm{C}}\). Now we can find the energy gained by the oil in the radiator: \(E_\text{oil} = 38 \times 2200 \times 40 = 3344000\,\mathrm{J}\).

Determine the Energy Gained by the Air in the Room

Similarly to step 2, we'll use the equation \(E_\text{air}=mc\Delta T\) to find the energy gained by the air in the room. The initial temperature of the air is \(10\,^{\circ}\mathrm{C}\), and the final temperature is \(20\,^{\circ}\mathrm{C}\). Therefore, \(\Delta T = 20 - 10 = 10\,^{\circ}\mathrm{C}\). The mass of air in the room can be found using the equation \(m=\rho V\), where \(\rho\) is the density of air and \(V\) is the volume of the room. Assuming the density of air is approximately \(1.2\,\mathrm{kg/m^3}\) and given that the volume of the room is \(50\,\mathrm{m^3}\), we find the mass of the air in the room to be \(m = 1.2 \times 50 = 60\,\mathrm{kg}\). The specific heat capacity of air, \(c\), is approximately \(1000\,\mathrm{J/kg\cdot^{\circ}\mathrm{C}}\). Now we can find the energy gained by the air in the room: \(E_\text{air} = 60 \times 1000 \times 10 = 600000\,\mathrm{J}\).

Determine the Energy Lost From the Room

Given that heat is lost from the room at an average rate of \(0.35\,\mathrm{kJ/s}\), which is equal to \(350\,\mathrm{J/s}\), we can find the total energy lost from the room during the time \(t\) as \(E_\text{lost} = 350t\,\mathrm{J}\).

Set Up and Solve the Energy Conservation Equation

Now we apply the energy conservation principle. The energy supplied by the radiator must be equal to the energy gained by the oil in the radiator, the energy gained by the air in the room, and the energy lost from the room: \(E_\text{supplied} = E_\text{oil} + E_\text{air} + E_\text{lost}\). Substituting the expressions from steps 1-4, we get \(2400t = 3344000 + 600000 + 350t\). Simplifying the equation, we get \(2050t = 3944000\). Solving for \(t\), we find that \(t = \dfrac{3944000}{2050} = 1924\) seconds. Converting seconds to minutes, we obtain \(t \approx 32\) minutes. Therefore, the heater is kept on for approximately 32 minutes.