Question

A 25 -cm-diameter, 2.4-m-long vertical cylinder containing ice at \(0^{\circ} \mathrm{C}\) is buried right under the ground. The cylinder is thin-shelled and is made of a high thermal conductivity material. The surface temperature and the thermal conductivity of the ground are \(18^{\circ} \mathrm{C}\) and \(0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) respectively. The rate of heat transfer to the cylinder is (a) \(37.2 \mathrm{~W}\) (b) \(63.2 \mathrm{~W}\) (c) \(158 \mathrm{~W}\) (d) \(480 \mathrm{~W}\) (e) \(1210 \mathrm{~W}\)

Short answer

Answer: The rate of heat transfer to the cylinder is 63.2 W.

Step-by-step solution

Calculate cylinder's surface area

Given a diameter of 25 cm, the half of it will be the radius (\(12.5\) cm). Change its unit to meters (m) by dividing it by \(100\). So, the radius is \(0.125\) m. We also have a cylinder length, \(L\), of \(2.4\) meters. The surface area of a cylinder A is calculated by: $$ A = 2\pi rL $$ $$ A = 2\pi (0.125)(2.4) $$ Calculate the area for further use in formula: $$ A \approx 0.9425 \mathrm{m}^2 $$

Calculate the temperature difference

The surface temperature of the ground is \(18^{\circ} \mathrm{C}\). The temperature of the ice inside the cylinder is \(0^{\circ} \mathrm{C}\). So the temperature difference \(\Delta T\) is: $$ \Delta T = 18^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C} = 18\mathrm{K} $$

Find the rate of heat transfer

Now that we have calculated the surface area and temperature difference, we can use the formula to find the rate of heat transfer, \(Q\). The thermal conductivity of the ground, \(k\), is given as \(0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). $$ Q = kA\frac{\Delta T}{L} $$ Inserting the values, the expression becomes: $$ Q = (0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}) \cdot (0.9425 \mathrm{m}^2) \frac{(18\mathrm{K})}{(2.4 \mathrm{m})} $$ Calculate the rate of heat transfer \(Q\): $$ Q \approx 63.2\,\mathrm{W} $$ In conclusion, the rate of heat transfer to the cylinder is \(63.2\, \mathrm{W}\) (option b).

Key concepts

Thermal Conductivity

Thermal conductivity is a material property that measures the ability of a material to conduct heat. It represents how easily heat passes through a material. The higher the thermal conductivity, the more efficient the material is at transferring heat. For example, metals typically have high thermal conductivity, making them good at releasing or absorbing heat quickly, which is why a metal spoon heats up fast when placed in a hot liquid.

In the context of the exercise, the thermal conductivity of the ground, denoted by the symbol \(k\), is given as \(0.85 \,\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\). This value is a constant for a specific material and is used to calculate the rate of heat transfer, which depends on the material’s ability to conduct heat. High thermal conductivity materials are chosen in applications where efficient heat transfer is needed, like in the thin shell of our vertical cylinder containing ice.

Temperature Difference

The temperature difference, often indicated as \(\Delta T\), plays an essential role in calculating the heat transfer rate. It is the driving force behind heat transfer in that heat moves from a region of higher temperature to a region of lower temperature. The greater the temperature difference between two regions, the faster heat will flow from the hot to the cold region.

In this exercise, the temperature difference is the difference between the surface temperature of the ground at \(18^\circ \mathrm{C}\) and the ice inside the cylinder at \(0^\circ \mathrm{C}\), which is \(18 \mathrm{K}\) when converted to Kelvin. Kelvin is used instead of Celsius in heat transfer calculations because it provides an absolute scale where \(0 \mathrm{K}\) represents absolute zero.

Cylinder Surface Area

The cylinder surface area is critical for calculating the heat transfer rate because it determines the size of the area over which heat can be exchanged. A larger surface area allows for more heat to be transferred. The formula for the surface area \(A\) of a cylinder is given by \[A = 2\pi rL\], where \(r\) is the radius of the base and \(L\) is the length of the cylinder.

In this scenario, with the given dimensions of the cylinder (a diameter of 25 cm and a length of 2.4 m), we first find the radius (half of the diameter) and then calculate the surface area. This surface area is then used in the heat transfer equation to determine the rate at which heat is transferred to the ice in the cylinder. A larger cylinder would have a bigger surface area, and therefore, all else being equal, a higher heat transfer rate.