Question

A \(100 .\) mm by \(100 .\) mm by 5.00 mm block of ice at \(0^{\circ} \mathrm{C}\) is placed on its flat face on a 10.0 -mm-thick metal disk that covers a pot of boiling water at normal atmospheric pressure. The time needed for the entire ice block to melt is measured to be \(0.400 \mathrm{~s} .\) The density of ice is \(920 . \mathrm{kg} / \mathrm{m}^{3} .\) Use the data in Table 18.3 to determine the metal the disk is most likely made of

Short answer

Answer: Copper

Step-by-step solution

Calculate the mass of the ice block

First, we need to determine the mass of the ice block. We can do this using its dimensions and the given density: \(\textrm{Mass} = \textrm{Volume} \times \textrm{Density}\) The volume of the ice block is \((0.1 \textrm{ m} \times 0.1 \textrm{ m} \times 0.005 \textrm{ m}) = 5.0 \times 10^{-5} \textrm{ m}^3\). The mass is then \((5.0 \times 10^{-5} \textrm{ m}^3)(920 \textrm{ kg/m}^3) = 4.6 \textrm{ kg}\).

Calculate the heat required to melt the ice block

Next, we need to determine the amount of heat needed to melt the entire ice block. We can use the formula \(Q = mL\), where \(Q\) is the amount of heat, \(m\) is the mass of the ice block, and \(L\) is the latent heat of fusion for ice (which is \(3.34 \times 10^5 \textrm{ J/kg}\)). \(Q = (4.6 \textrm{ kg})(3.34 \times 10^5 \textrm{ J/kg}) = 1.53 \times 10^6 \textrm{ J}\)

Calculate the rate of heat transfer

Given the time it takes for the ice block to melt, we can determine the rate at which heat is transferred from the metal disk to the ice block: \(\textrm{Rate of heat transfer} = \dfrac{\textrm{Heat}}{\textrm{Time}} = \dfrac{1.53 \times 10^6 \textrm{ J}}{0.400 \textrm{ s}} = 3.83 \times 10^6 \textrm{ J/s}\)

Use the formula for heat transfer through conduction

Now we can use the formula for heat conduction to relate the rate of heat transfer to the characteristics of the metal disk: \( \dfrac{Q}{t} = k A \dfrac{T_{2} - T_{1}}{d} \) where \(k\) is the thermal conductivity of the metal, \(A\) is the area of contact between the metal disk and the ice block, \(T_{1}\) and \(T_{2}\) are the initial and final temperatures, and \(d\) is the thickness of the metal disk. Rearrange the formula to solve for the thermal conductivity \(k\): \(k = \dfrac{\dfrac{Q}{t} \times d}{A (T_{2} - T_{1})}\)

Calculate the thermal conductivity of the metal

We have enough information to calculate the thermal conductivity of the metal. The area of contact \(A\) is \(0.1 \textrm{ m} \times 0.1 \textrm{ m} = 0.01 \textrm{ m}^2\), the temperature difference between the metal disk and the ice block is \((100 - 0) \textrm{ C}\), and the thickness of the metal disk is \(0.01 \textrm{ m}\). Plug these values into the formula: \(k = \dfrac{(3.83 \times 10^6 \textrm{ J/s}) \times 0.01 \textrm{ m}}{0.01 \textrm{ m}^2 \times 100 \textrm{ C}} = 383 \textrm{ W/mK}\)

Compare with the data in Table 18.3

Finally, compare the calculated thermal conductivity with the values in Table 18.3 to determine which metal the disk is most likely made of. The value of \(383 \textrm{ W/mK}\) is closest to the thermal conductivity of copper (\(398 \textrm{ W/mK}\)), suggesting that the metal disk is most likely made of copper.

Key concepts

Understanding Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another due to a temperature difference. There are three main modes of heat transfer: conduction, convection, and radiation. When studying the melting of an ice block, as in the problem presented, conduction is the key mode of heat transfer involved.
In the exercise, heat transfer occurs from the metal disk, which is being heated by the pot of boiling water, to the ice block that is in contact with the metal. The rate of heat transfer, which is energy per unit time, is critical for determining how quickly the ice block melts. Understanding heat transfer is vital to solving problems in thermodynamics, material science, and engineering. By being able to take a quantity like the amount of heat required to melt ice and then relate it to time, students grasp how thermal energy moves and transforms in real-world scenarios.

Latent Heat of Fusion Explained

The latent heat of fusion is the amount of heat required to change a substance from the solid phase to the liquid phase at constant temperature and pressure. It is a type of latent heat, the energy released or absorbed during a phase change, without a change in temperature of the substance.
In our textbook example, we calculate the amount of energy needed to melt the ice block using the latent heat of fusion of ice. This concept is pivotal because it defines the thermal energy required to overcome the molecular forces that hold the solid structure together, resulting in a phase change to liquid. Interestingly, the amount of heat necessary for the phase change does not change the temperature of the ice; it only changes its state. This property is what makes calculating the energy transfer during phase changes such as melting or freezing different from simply changing the temperature of a substance.

Conduction as a Heat Transfer Mechanism

Conduction is the transfer of heat through a material without the bulk movement of the substance. It occurs due to the collision of molecules within a body at different temperatures. The rate at which conduction happens depends on several factors: the temperature gradient, the cross-sectional area through which heat is travelling, the material's thickness, and a very important property known as thermal conductivity.
Thermal conductivity, represented by 'k' in formulas, is a measure of a material's ability to conduct heat. It varies from material to material and influences how quickly heat can move through a substance. In the solved problem, we see how crucial thermal conductivity is in determining the rate of heat transfer from the metal disk to the ice. The calculation of 'k' allowed us to deduce the metal type based on its proximity to known values for various metals, indicating the disk is likely made of copper. Understanding conduction and thermal conductivity helps students tackle a range of problems, from designing heat exchangers to managing heat dissipation in electronic devices.