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: \(4.50 \mathrm{~min}\)

Short answer

Answer: The time required for the milk to warm up is approximately 4.50 minutes.

Step-by-step solution

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: \(m = \rho V = 1000\mathrm{~kg/m^3} \frac{63\pi\mathrm{~cm^3}}{10^6\mathrm{~m^3/cm^3}} = 0.063\pi\mathrm{~kg}\).

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. \(\Delta T = T_{final} - T_{initial} = 38^{\circ}\mathrm{C} - 3^{\circ}\mathrm{C} = 35\mathrm{~K}\) \(Q = (0.063\pi\mathrm{~kg})(4.2\mathrm{~kJ/kg\cdot K})(35\mathrm{~K}) = 9.282\pi\mathrm{~kJ}\)

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: \(q = (120\mathrm{~W/m^2\cdot K})(0.006\pi\mathrm{~m^2})(32\mathrm{~K}) = 72\pi\mathrm{~W}\)

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 \(4.50\mathrm{~min}\).

Key concepts

Heat Transfer Coefficient

The heat transfer coefficient is a crucial element in heat transfer problems. It represents the ease with which heat is exchanged between two materials. In this scenario, we look at the heat transfer between hot water and a cylindrical container filled with milk. The coefficient is given as \(120\, \text{W/m}^2\cdot \text{K}\).
This means for every square meter of surface area, and for every degree Kelvin of temperature difference between these materials, 120 watts of heat will be transferred. A higher coefficient would imply a more effective heat transfer.
The heat transfer coefficient is essential in calculating how quickly or slowly an object will heat up or cool down. Its value depends on several factors such as the type of fluid flow, the roughness of the surfaces in contact, and the specific surface area engaged in the transfer. It connects to real-world contexts, such as engineering designs for heating systems to ensure energy efficiency.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius or Kelvin. In our exercise, we use the specific heat capacity of water, \(c_p = 4.2\,\text{kJ/kg}\cdot \text{K}\). It's this property that determines how much energy is needed to change the temperature of milk from \(3^{\circ} \text{C}\) to \(38^{\circ} \text{C}\).
This property plays a critical role in calculations involving temperature changes, as it influences how substances react to heat transfer. Substances with a high specific heat capacity can absorb more heat before their temperature rises significantly. This trait is essential, for example, in applications like thermal buffers or insulation materials, where heat absorption without large temperature changes is desired.
Understanding specific heat helps in various fields, from designing climate control systems to improving culinary techniques where precise temperature control is vital.

Calculating Heat Transfer

The total heat transfer required, \(Q\), can be calculated using the formula \(Q = mc_p\Delta T\). Here, \(m\) represents the mass of the milk, \(c_p\) the specific heat capacity, and \(\Delta T\) the temperature change.
In this example, converting initial energy units and using the height and diameter of the container, we compute the required heat to change milk's temperature by \(35\, \text{K}\).
Critical calculations draw on both the specific heat capacity and mass, showing their interdependence. This understanding allows us to predict how much energy is necessary to heat or cool substances. These predictions are vital in different sectors, such as manufacturing, where precise energy usage monitoring can cut costs and improve efficiency.
Overall, calculating heat transfer is a foundational skill in thermodynamics, providing insights into energy consumption and the efficiency of heating or cooling processes.

Cylindrical Container

Cylindrical containers have unique geometric properties that influence heat transfer calculations. The container's dimensions, such as diameter and height, determine its surface area and volume, directly affecting how heat is transferred to its contents.
In our example, the thin-walled cylindrical container fully submerged in hot water allows us to consider all surfaces in contact with the heat source. This approach mirrors many real-world scenarios where containers are placed in heat baths, making understanding crucial for everyday applications such as cooking and industrial processing.
By knowing the formula for the surface area, \(A = 2\pi rh + 2\pi r^2\), students can accurately compute the contact area where heat exchange occurs. This computation is vital in scenarios like designing reactors or storage tanks where maximizing or minimizing heat transfer is important.
Understanding how geometric factors of containers influence thermal dynamics enhances many fields, from engineering to everyday household applications.