Question

Mechanically polished, 5 -cm-diameter, stainless steel ball bearings are heated to \(125^{\circ} \mathrm{C}\) uniformly. The ball bearings are then submerged in water at \(1 \mathrm{~atm}\) to be cooled. Determine the rate of heat that is removed from a ball bearing at the instant it is submerged in the water.

Short answer

Answer: The approximate rate of heat transfer from the hot stainless steel ball bearing to the cool water at the initial instant is 196.47 Watts.

Step-by-step solution

Find the heat transfer coefficient between the ball bearing and water

We are not given the heat transfer coefficient between the ball bearing and the water, but we can estimate it. For a typical range of fluid velocities and temperatures, the heat transfer coefficient for forced convection between a solid object and water lies between 10 and 10,000 \(W/m^2-K\). Since we have no additional information about the system, let's assume a moderate value of 1000 \(W/m^2-K\) for our calculations. This will allow us to calculate an approximate rate of heat removal.

Calculate the surface area of the ball bearing

The surface area of a sphere can be calculated using the formula: \(A = 4 \pi r^2\) Given the diameter of the ball bearing is 5 cm, the radius is 2.5 cm or 0.025 m. Substituting the radius value in the formula, we get: \(A = 4 \pi (0.025^2) = 0.00785 \; m^2\)

Convert the temperatures from Celsius to Kelvin

Newton's Law of Cooling deals with temperature differences; therefore, we should convert the temperatures to Kelvin. The conversion formula is: \(T_{Kelvin} = T_{Celsius} + 273.15\) For the initial temperature of the ball bearing, \(T_i = 125^{\circ} C + 273.15 = 398.15 K\) Standard water temperature at atmospheric pressure is 100 degrees Celsius, so \(T_{water} = 100^{\circ} C + 273.15 = 373.15 K\)

Apply Newton's Law of Cooling to calculate the rate of heat transfer

The formula for Newton's Law of Cooling is: \(q = hA \Delta T\) Where \(q\) is the rate of heat transfer, \(h\) is the heat transfer coefficient, \(A\) is the surface area, and \(\Delta T\) is the temperature difference between the two objects (in this case, the ball bearing and the water). Now, we can plug in our values into the formula: \(q = 1000 (0.00785) (398.15 - 373.15)\) \(q = 196.47 W\)

Interpret the results

The rate of heat removed from the ball bearing at the instant it is submerged in the water is approximately 196.47 Watts. Note that our value for the heat transfer coefficient was an estimate, and the actual value could vary between 10 and 10,000 \(W/m^2-K\). Therefore, this rate of heat transfer is an approximation and could be higher or lower depending on the actual setup.