Question

An electric resistance space heater is designed such that it resembles a rectangular box \(50 \mathrm{~cm}\) high, \(80 \mathrm{~cm}\) long, and $15 \mathrm{~cm}\( wide filled with \)45 \mathrm{~kg}$ of oil. The heater is to be placed against a wall, and thus heat transfer from its back surface is negligible. The surface temperature of the heater is not to exceed $75^{\circ} \mathrm{C}\( in a room at \)25^{\circ} \mathrm{C}$ for safety considerations. Disregarding heat transfer from the bottom and top surfaces of the heater in anticipation that the top surface will be used as a shelf, determine the power rating of the heater in W. Take the emissivity of the outer surface of the heater to be \(0.8\) and the average temperature of the ceiling and wall surfaces to be the same as the room air temperature. Also, determine how long it will take for the heater to reach steady operation when it is first turned on (i.e., for the oil temperature to rise from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\) ). State your assumptions in the calculations.

Short answer

Step 1: Calculate the surface area of the heater. A = (0.50 m) x (0.80 m) = 0.4 m² Step 2: Calculate the power rating using the Stefan-Boltzmann Law. q = εσA(Ts^4 - Tr^4) = 0.8 × 5.67 × 10^(-8) × 0.4 × (348^4 - 298^4) ≈ 427.5 W Step 3: Calculate the energy required to raise the oil temperature. E = mcΔT = 45 kg × 2.0 × 10^3 J/kg.K × 50 K = 4.5 × 10^6 J Step 4: Calculate the time taken to reach steady-state operation. t = E/q = (4.5 × 10^6 J) / (427.5 W) ≈ 10526.4 s, which is approximately 175.4 minutes. Answer: The power rating of the space heater is 427.5 W, and it takes approximately 175.4 minutes to reach steady-state operation.

Step-by-step solution

Calculate the surface area of the heater

Since heat transfer from the back, bottom, and top surface is negligible, we only need the surface area of the front surface. The front surface has dimensions \(50 \mathrm{~cm}\) in height and \(80 \mathrm{~cm}\) in length. Convert these dimensions to meters and multiply them to get the surface area: \[ A = (0.50 \,\text{m})\times(0.80\, \text{m}) \]

Calculate the power rating using the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the heat transfer rate through radiation can be calculated as follows: \[ q = \epsilon \sigma A(T_s^4 - T_r^4) \] Here, \(\epsilon = 0.8\) is the emissivity, \(\sigma = 5.67\times10^{-8} \,\text{W/m}^2\text{K}^4\) is the Stefan-Boltzmann constant, \(T_s = 75^{\circ}\text{C} = 348\,\text{K}\) is the surface temperature, and \(T_r = 25^{\circ}\text{C} = 298\,\text{K}\) is the room air temperature. Plug in all the values and solve for the heat transfer rate \(q\), which is equal to the power rating of the heater.

Calculate the energy required to raise the oil temperature

To determine the energy required to raise the oil temperature from \(25^{\circ}\text{C}\) to \(75^{\circ}\text{C}\), we use the following equation: \[ E = mc\Delta T \] where \(m = 45\,\text{kg}\) is the mass of oil, and \(c\) is the specific heat capacity of the oil. Assuming the oil is similar to a common mineral oil, the specific heat capacity is approximately \(c = 2.0 \times 10^3 \,\text{J/kg}\cdot\text{K}\). Calculate the energy \(E\) using the given values and the temperature difference \(\Delta T = 50\,\text{K}\).

Calculate the time taken to reach steady-state operation

Since the power rating of the heater (the heat transfer rate) is known, we can calculate the time taken to reach steady-state operation: \[ t = \frac{E}{q} \] where \(t\) is the time taken in seconds. Substitute the calculated energy \(E\) and power rating \(q\) to obtain the time taken to reach steady-state operation. You may want to convert the result to minutes for convenience.