To warm up some milk for a baby, a mother pours milk into a thin-walled cylindrical container whose diameter is \(6 \mathrm{~cm}\). The height of the milk in the container is \(7 \mathrm{~cm}\). She then places the container into a large pan filled with hot water at \(70^{\circ} \mathrm{C}\). The milk is stirred constantly, so that its temperature is uniform at all times. If the heat transfer coefficient between the water and the container is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the milk to warm up from \(3^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\). Assume the entire surface area of the cylindrical container (including the top and bottom) is in thermal contact with the hot water. Take the properties of the milk to be the same as those of water. Can the milk in this case be treated as a lumped system? Why? Answer: \(4.50 \mathrm{~min}\)

Step by step solution

01

Determine the volume and mass of the milk

First, we need to find the volume of the milk inside the container. The container is a cylinder with diameter \(6\mathrm{~cm}\) and height \(7\mathrm{~cm}\). We will find the volume using the formula: \(V = \pi r^2h\), where \(r\) is the radius of the cylinder and \(h\) is its height. The radius is half of the diameter, so in this case, it is \(3\mathrm{~cm}\). Therefore, the volume of the milk is: \(V = \pi (3\mathrm{~cm})^2 (7\mathrm{~cm}) = 63\pi\mathrm{~cm}^3\). Assuming the properties of the milk are the same as water, we can use the density of water \(\rho = 1000\mathrm{~kg/m^3}\) to find the mass of the milk: \(m = \rho V = 1000\mathrm{~kg/m^3} \frac{63\pi\mathrm{~cm^3}}{10^6\mathrm{~m^3/cm^3}} = 0.063\pi\mathrm{~kg}\).

02

Find the total heat transfer required to heat the milk

We know the initial and final temperatures of the milk, so we can find the heat transfer required to warm the milk using the specific heat capacity \(c_p\) of water (\(c_p = 4.2\mathrm{~kJ/kg\cdot K}\)) and the equation \(Q = mc_p\Delta T\), where \(\Delta T\) is the change in temperature. \(\Delta T = T_{final} - T_{initial} = 38^{\circ}\mathrm{C} - 3^{\circ}\mathrm{C} = 35\mathrm{~K}\) \(Q = (0.063\pi\mathrm{~kg})(4.2\mathrm{~kJ/kg\cdot K})(35\mathrm{~K}) = 9.282\pi\mathrm{~kJ}\)

03

Compute the heat transfer between water and the container

The heat transfer coefficient \(h\) is given as \(120\mathrm{~W/m^2\cdot K}\). We need to find the total surface area of the container, including the top and bottom. The formula for the surface area of a cylinder is \(A = 2\pi rh + 2\pi r^2\), where \(r\) is the radius and \(h\) is the height. Using the given dimensions, the surface area is: \(A = 2\pi(3\mathrm{~cm})(7\mathrm{~cm}) + 2\pi (3\mathrm{~cm})^2 = 42\pi + 18\pi\mathrm{~cm^2}\) We need to convert the surface area to square meters: \(A = (42\pi + 18\pi\mathrm{~cm^2})\frac{1\mathrm{~m^2}}{10^4\mathrm{~cm^2}} = 0.006\pi\mathrm{~m^2}\) The temperature difference between the water and the milk is constant at \(70^{\circ}\mathrm{C} - 38^{\circ}\mathrm{C} = 32\mathrm{~K}\). Using the heat transfer formula, \(q = hA\Delta T\), we have: \(q = (120\mathrm{~W/m^2\cdot K})(0.006\pi\mathrm{~m^2})(32\mathrm{~K}) = 72\pi\mathrm{~W}\)

04

Determine the time required for the milk to heat up

Now that we have the heat transfer rate between the water and the container, we can use the equation \(t = \frac{Q}{q}\) to find the time required for the milk to warm up from \(3^{\circ}\mathrm{C}\) to \(38^{\circ}\mathrm{C}\). \(t = \frac{9.282\pi\mathrm{~kJ}}{72\pi\mathrm{~W}} = \frac{9.282\pi\mathrm{~kJ}\,\cdot 1000\mathrm{~J/kJ}}{72\pi\mathrm{~W}} = 129\mathrm{~s}\) The time required for the milk to warm up is \(129\mathrm{~s}\), which is approximately \(4.50\mathrm{~min}\).

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