Question

In a manufacturing facility, 5 -cm-diameter brass balls \(\left(\rho=8522 \mathrm{kg} / \mathrm{m}^{3} \text { and } c_{p}=0.385 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) initially at \(120^{\circ} \mathrm{C}\) are quenched in a water bath at \(50^{\circ} \mathrm{C}\) for a period of \(2 \mathrm{min}\) at a rate of 100 balls per minute. If the temperature of the balls after quenching is \(74^{\circ} \mathrm{C}\), determine the rate at which heat needs to be removed from the water in order to keep its temperature constant at \(50^{\circ} \mathrm{C}\).

Short answer

Answer: The rate of heat removal required is 990.6 kW.

Step-by-step solution

Find the mass of one brass ball

Since we are given the diameter and density of the brass balls, first we should find the volume of one ball, then use the volume to determine its mass using the formula \(m = \rho V\). The volume of a sphere can be calculated by the formula \(V = \frac{4}{3}\pi r^3\) where r is the radius of the sphere. The diameter of a ball is 5 cm, so the radius can be found by dividing the diameter by 2: \(r = \frac{d}{2} = \frac{5\,\text{cm}}{2} = 2.5\,\text{cm} = 0.025\,\text{m}\) Now, we can find the volume: \(V =\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.025\,\text{m})^3 = 6.544 \times 10^{-5}\,\text{m}^{3}\) Now, we can calculate the mass of a ball: \(m = \rho V = 8522\,\text{kg/m}^3 \times 6.544 \times 10^{-5}\,\text{m}^{3} = 0.5577\,\text{kg}\)

Determine the temperature difference

The temperature difference between the initial and final state of the brass balls can be found by subtracting their final temperature (74°C) from their initial temperature (120°C): \(\Delta T = T_\text{initial} - T_\text{final} = 120^{\circ}\mathrm{C} - 74^{\circ}\mathrm{C}= 46^{\circ}\mathrm{C}\)

Calculate heat transfer

Now we have all the information necessary to find the amount of heat transferred from one brass ball to the water. The formula for heat transfer is \(Q = mc_p\Delta T\), where \(m\) is the mass of the brass ball, \(c_p\) is its specific heat, and \(\Delta T\) is the temperature difference: \(Q = mc_p\Delta T = 0.5577\,\text{kg} \times 0.385\,\text{kW\cdot min/kg \cdot C} \times 46^{\circ} \mathrm{C} = 9.906\,\text{kW\cdot min}\)

Calculate the rate of heat removal

We are given that the rate of quenching is 100 balls per minute and the quenching time is 2 minutes. Therefore, the total number of balls quenched in 2 minutes is 200. The total heat transferred from all balls can be calculated by multiplying the heat transfer of one ball by the total number of balls: \(Q_\text{total} = Q \times \text{number of balls} = 9.906\,\text{kW\cdot min} \times 200 = 1981.2\,\text{kW\cdot min}\) Since we need to find the rate of heat removal, we need to divide the total heat transfer by the total quenching time of 2 minutes: Heat removal rate = \(\frac{Q_\text{total}}{\text{time}} = \frac{1981.2\,\text{kW\cdot min}}{2\,\text{min}} = 990.6\,\text{kW}\) Therefore, the rate at which heat needs to be removed from the water in order to keep its temperature constant at \(50^{\circ} \mathrm{C}\) is \(990.6\,\text{kW}\).