Question

A 5.00 -g sample of aluminum pellets (specific heat capacity \(=\) 0.89 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity \(=0.45 \mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) are heated to \(100.0^{\circ} \mathrm{C}\) . The mixture of hot iron and aluminum is then dropped into 97.3 \(\mathrm{g}\) water at \(22.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Short answer

The final temperature of the metal and water mixture is approximately 25.6°C.

Step-by-step solution

Identify the known values and the unknown final temperature

We are given the following values: - Mass of aluminum, m_aluminum = 5.00 g - Specific heat capacity of aluminum, c_aluminum = 0.89 J/g°C - Mass of iron, m_iron = 10.00 g - Specific heat capacity of iron, c_iron = 0.45 J/g°C - Initial temperature of both aluminum and iron = 100°C - Mass of water, m_water = 97.3 g - Specific heat capacity of water, c_water = 4.18 J/g°C (default value) - Initial temperature of water = 22.0°C The final temperature of the system (water and metals) is unknown. We will call it T_final.

Calculate the heat gained by the water

First, we need to calculate the amount of heat gained by the water when it is heated to the final temperature. We can use the formula: q_water = m_water × c_water × (T_final - initial_temperature_water) Note that q_water is positive because it represents the heat gained by the water.

Calculate the heat lost by aluminum and iron

Next, we need to calculate the heat lost by the aluminum and iron as they cool down to the final temperature. We can do this for each metal using the formula: q_aluminum = m_aluminum × c_aluminum × (initial_temperature_metal - T_final) q_iron = m_iron × c_iron × (initial_temperature_metal - T_final) Note that q_aluminum and q_iron are both negative because they represent the heat lost by the metals.

Apply conservation of energy

Since there is no heat loss to the surroundings, the heat gained by the water must be equal to the heat lost by the metals. Therefore: q_water = q_aluminum + q_iron Now, substitute the values from Steps 2 and 3 and solve for T_final: m_water × c_water × (T_final - initial_temperature_water) = m_aluminum × c_aluminum × (initial_temperature_metal - T_final) + m_iron × c_iron × (initial_temperature_metal - T_final)

Solve for T_final

By plugging in the given values, we can now solve for T_final: 97.3 \times 4.18 \times (T_final - 22) = 5.00 \times 0.89 \times (100 - T_final) + 10.00 \times 0.45 \times (100 - T_final) Solving this equation gives us the final temperature: T_final ≈ 25.6°C The final temperature of the metal and water mixture is approximately 25.6°C.

Key concepts

Specific Heat Capacity

When it comes to understanding heat transfer, specific heat capacity is a crucial concept. It tells us how much energy is needed to raise the temperature of a substance by one degree Celsius for one gram of the substance. In our exercise, both the aluminum and iron pellets have different specific heat capacities: 0.89 J/g°C for aluminum and 0.45 J/g°C for iron.

This means that aluminum requires more energy to change its temperature compared to iron. Why is that? Well, specific heat capacity depends on the material's structure. Materials with higher specific heat can store more energy, while those with lower values change temperature more easily with less energy input. This is why, in thermal experiments like these, understanding specific heat is essential for predicting how substances will react when mixed or heated. By knowing these values, we can accurately calculate how much energy (heat) will be gained or lost by each substance as they reach the same final temperature.

Conservation of Energy

At the core of this thermal equilibrium problem is the principle of conservation of energy. This law states that energy cannot be created or destroyed, only transferred or transformed.

In this exercise, we have no heat loss to the surroundings, which perfectly illustrates this principle. When hot metal pellets are mixed with cooler water, heat transfers from the metals to the water until they reach a thermal equilibrium. The equation we use, \[ q_{\text{water}} = q_{\text{aluminum}} + q_{\text{iron}} \]ensures that the energy balance is maintained. The heat gained by the water comes from the heat lost by aluminum and iron. By setting these quantities equal, we can solve for the final temperature.In practice, this concept helps us not only in theoretical calculations but also in designing practical applications such as heating systems and thermal management in various fields. It's detailed understanding that aids engineers and scientists in creating efficient energy solutions.

Metal Heat Transfer

Metal heat transfer is an interesting and complex topic. In the given exercise, two different metals, aluminum and iron, are used. These metals start at a higher temperature than the water they are mixed with, at 100°C.

The metals, because they are at a higher temperature, will start to transfer heat to the cooler water. Metals are good conductors of heat, which means they can transfer energy quickly. This is why metals are often used in situations where efficient heat transfer is needed, like in cookware or radiators. In our exercise, the process of these metals reaching the same temperature as the water is due to their ability to transfer heat efficiently. The specific heat capacity and the mass of these metals determine how quickly and how much heat is transferred. With this process, the thermal equilibrium is achieved and we observe a final uniform temperature of about 25.6°C. Understanding how metals transfer heat helps us create and optimize many technologies that involve thermal processes, from engines to electronic devices, where managing heat effectively is crucial.