Question

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: \(4.50\) min

Short answer

Answer: It takes about 8.86 minutes to heat the milk from 3°C to 38°C, and the milk cannot be treated as a lumped system in this case.

Step-by-step solution

Calculate the volume of the milk

The volume (V) of the cylindrical container filled with milk is given by the formula: $$V = \frac{\pi d^2}{4} \times h$$ Let's convert the diameter and height to meters and then calculate the volume. $$V = \frac{\pi (0.06 \mathrm{~m})^2}{4} \times 0.07 \mathrm{~m} = 0.0001885 \mathrm{~m}^3$$

Find the mass of the milk

Using the density of the milk (which is assumed to be the same as water), we can calculate the mass of the milk as follows: $$m = \rho \times V = (1000 \mathrm{~kg} / \mathrm{m}^3) \times 0.0001885 \mathrm{~m}^3 = 0.1885 \mathrm{~kg}$$

Calculate the surface area of the container

The surface area (A) of the cylindrical container in contact with the water (including the top and bottom) can be calculated as: $$A = \pi d^2 + 2 \pi \left(\frac{d}{2} \right)h = 2 \pi \left(\frac{d}{2} \right) \times (d + 2h)$$ $$A = 2 \pi \left(0.03 \mathrm{~m} \right) \times (0.06 \mathrm{~m} + 2 \times 0.07 \mathrm{~m}) = 0.02568 \mathrm{~m}^2$$

Determine the time required for the milk to reach the final temperature

The total heat (Q) required to raise the milk's temperature from \(T_i\) to \(T_f\) can be calculated using the formula: $$Q = m \times c_p \times \Delta T = m \times c_p \times (T_f - T_i)$$ $$Q = (0.1885 \mathrm{~kg}) \times (4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}) \times (38 - 3)^{\circ} \mathrm{C} = 27438 \mathrm{~J}$$ The heat transfer rate (Q') can be calculated using the formula: $$Q' = h \times A \times (T_w - T_m)$$ Here, \(T_w\) is the temperature of the water and \(T_m\) is the average temperature of the milk. We will take \(T_m\) as the midpoint of \(T_i\) and \(T_f\). $$Q' = (120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}) \times 0.02568 \mathrm{~m}^2 \times (70 - \frac{3 + 38}{2})^{\circ} \mathrm{C} = 51.61 \mathrm{~W}$$ Now, we can find the time (t) required for the milk to reach the final temperature: $$t = \frac{Q}{Q'} = \frac{27438 \mathrm{~J}}{51.61 \mathrm{~W}} = 531.48 \mathrm{~s}$$ $$t = 531.48 \mathrm{~s} = 8.86 \mathrm{~min}$$

Check if the milk can be treated as a lumped system

To check if the milk can be treated as a lumped system, we calculate the Biot number (Bi): $$Bi = \frac{hL_c}{k}$$ Here, \(L_c\) is the characteristic length, which can be calculated as: $$L_c = \frac{V}{A}$$ $$L_c = \frac{0.0001885 \mathrm{~m}^3}{0.02568 \mathrm{~m}^2} = 0.007339\ \mathrm{m}$$ Now, we calculate the Biot number: $$Bi = \frac{(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}) \times 0.007339 \mathrm{~m}}{0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 1.467$$ Since the Biot number \(Bi > 0.1\), the milk cannot be treated as a lumped system. In conclusion, it takes about 8.86 minutes to heat the milk from \(3^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) and the milk cannot be treated as a lumped system in this case.