Question

A metal brick found in an excavation was sent to a testing lab for nondestructive identification. The lab weighed the sample brick and found its mass to be \(3.0 \mathrm{~kg} .\) The brick was heated to a temperature of \(3.0 \cdot 10^{2}{ }^{\circ} \mathrm{C}\) and dropped into an insulated copper calorimeter of mass 1.5 kg containing \(2.0 \mathrm{~kg}\) of water at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} .\) The final temperature at equilibrium was noted to be \(31.7^{\circ} \mathrm{C}\). By calculating the specific heat of the sample from this data, can you identify the brick's material?

Short answer

Answer: To calculate the specific heat capacity of the metal brick, use the following equation: \(c_b=\frac{m_w c_w (T_f-T_{wi})}{m_b (T_{f}-T_{bi})}\), where \(m_w\) and \(m_b\) denote the masses of water and brick, \(c_w\) is the specific heat capacity of water, and \(T_{wi}\), \(T_{bi}\), and \(T_f\) are the initial temperatures of water and brick and the final equilibrium temperature, respectively. Once the specific heat capacity of the brick, \(c_b\), is found, it can be compared to the specific heat capacities of common metals to possibly identify the material.

Step-by-step solution

Calculate the heat gained by the water

Since we know the mass of water \(m_w = 2.0 kg\), the initial temperature \(T_{wi}=20^\circ\,C\), final temperature \(T_{f}=31.7^\circ\,C\), and the specific heat capacity of water \(c_w=4186\, J/(kg\cdot K)\), we can calculate the heat gained by water using the formula: \(Q_w=m_w c_w (T_f-T_{wi})\)

Calculate the heat lost by the brick

We have the mass of the metal brick \(m_b = 3.0\,kg\), initial temperature \(T_{bi}=300^\circ\,C\), and the final temperature \(T_{f}=31.7^\circ\,C\). Let the specific heat of the brick be \(c_b\). Therefore, the heat lost by the brick can be calculated as: \(Q_b=m_b c_b (T_{f}-T_{bi})\)

Equate the heat gained by water and heat lost by the brick

For the system to reach equilibrium, the heat gained by water will be equal to the heat lost by brick i.e., \(Q_w=Q_b\). Plugging in the values from steps 1 and 2, we get: \(m_w c_w (T_f-T_{wi}) = m_b c_b (T_{f}-T_{bi})\)

Calculate the specific heat of the brick

Solve for \(c_b\) in the equation above: \(c_b=\frac{m_w c_w (T_f-T_{wi})}{m_b (T_{f}-T_{bi})}\) Now, plug in the known values and solve for \(c_b\).

Identify the material of the brick

Once we find the specific heat capacity \(c_b\) of the brick, we can compare it to the specific heat capacities of common metals and try to identify the material. Note, however, that without a proper list of specific heat capacities of different materials or metals, we won't be able to identify the material directly. The found value may be used as an indication of which material it might be.

Key concepts

Thermal Equilibrium

Thermal equilibrium occurs when two objects that are in contact with each other reach the same temperature. In this state, no net heat exchange occurs as they have achieved balanced energy distribution. This concept is crucial in problems involving calorimetry, as it allows us to set the heat gained by one body equal to the heat lost by another body.
In the exercise, thermal equilibrium is reached when the heated metal brick is placed in the water inside the calorimeter. Over time, the brick loses heat, while the water absorbs it until both reach a common final temperature of 31.7°C.
This shared temperature indicates that thermal equilibrium has been achieved. Understanding this concept helps in calculating other values, such as the specific heat capacity of the unknown material.

Calorimetry

Calorimetry is a method used to measure the amount of heat involved in a chemical reaction, physical change, or heat capacity. By employing a calorimeter, we can accurately gauge how much heat is transferred between substances.
In the given problem, the calorimeter is an insulated container that helps track the heat transfer between the metal brick and the water. Because it is insulated, it ensures minimal heat exchange with the surrounding environment, keeping the focus on measuring the heat flow between the water and the brick.
The principle behind calorimetry in this context is that the heat lost by the brick (a decrease in its heat content) will be equal to the heat gained by the water. This relationship enables us to calculate the unknown specific heat capacity of the brick by comparing these heat exchanges.

Heat Transfer

Heat transfer is the process through which thermal energy moves from a hotter object to a colder one. It is one of the fundamental concepts examined in thermodynamics and plays a pivotal role in calorimetry experiments.
In the scenario described in the exercise, heat transfer occurs from the high-temperature metal brick to the water that is initially cooler.
There are three primary methods of heat transfer: conduction, convection, and radiation. In calorimetry, conduction is predominantly at play, where heat is transferred directly through the contact between the brick and the water.
Understanding how heat is transferred allows us to determine that as the brick's temperature drops, the water's temperature rises until they reach thermal equilibrium.

Material Identification

Material identification through calorimetry involves comparing the experimentally determined specific heat capacity of a sample with known values for different materials. This process is essential in identifying unknown samples, much like the problem exercise.
By using the formula obtained in the solution for specific heat capacity, students can calculate the specific heat of the brick. The formula is \[c_b=\frac{m_w c_w (T_f-T_{wi})}{m_b (T_{f}-T_{bi})}\] This calculation uses data from the experiment, such as the masses, temperatures, and known specific heat of water.
After finding this specific heat capacity, it is compared to a reference list of specific heats for various materials. This comparison helps in deducing the most likely candidate for the material of the brick.
However, it is important to note that in practical scenarios, precise identification can be challenging without additional context or data.