Question

Brass has a density of \(8.40 \mathrm{g} / \mathrm{cm}^{3}\) and a specific heat of \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} . \mathrm{A} 15.2 \mathrm{cm}^{3}\) piece of brass at an initial temperature of \(163^{\circ} \mathrm{C}\) is dropped into an insulated container with \(150.0 \mathrm{g}\) water initially at \(22.4^{\circ} \mathrm{C}\) What will be the final temperature of the brass-water mixture?

Short answer

The final temperature of the brass-water mixture is reached by equating the heat lost by the brass to the heat gained by the water and solving for the temperature. The exact value needs calculations based on the formulated equation and provided parameters.

Step-by-step solution

Identify given parameters

We are given the following parameters: brass density is \(8.40 \mathrm{g} / \mathrm{cm}^{3}\), brass specific heat is \(0.385 \mathrm{Jg}^{-1} \mathrm{C}^{-1}\), volume of the brass piece is \(15.2 \mathrm{cm}^{3}\), its initial temperature is \(163^{\circ}\mathrm{C}\), the mass of water is \(150.0 \mathrm{g}\), and its initial temperature is \(22.4^{\circ} \mathrm{C}\)

Calculate the loss of heat from the brass

First, find the mass of the brass using the formula \( \text{density} = \text{mass} / \text{volume} \). So, \( \text{mass} = \text{density} \times \text{volume} = 8.40 \mathrm{g} / \mathrm{cm}^{3} \times 15.2 \mathrm{cm}^{3}\) then using the formula for heat exchange \( Q = mcΔT \), we can calculate the heat loss of the brass. Here, \(m\) is mass, \(c\) is specific heat and \(ΔT\) is change in temperature, which is \(163 - T_{final}\)

Calculate the gain of heat by water

Applying the same formula \( Q = mcΔT \) on water, its heat gain is \(150.0 \mathrm{g} \times 4.18 \mathrm{Jg}^{-1} \mathrm{C}^{-1} \times (T_{final}-22.4^{\circ}\mathrm{C})\). Note, we've used the specific heat of water, which is \(4.18 \mathrm{Jg}^{-1} \mathrm{C}^{-1}\)

Solve for final temperature

Since the heat gained by the water is equal to the heat lost by the brass, we can equate the two equations from step 2 and 3 and solve for \(T_{final}\)

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics that describes how thermal energy moves from one entity to another. In the given exercise, brass and water are interacting in an isolated system, indicating no heat is exchanged with the external environment.

Understanding heat transfer is essential for grasping this exercise:

• Heat Loss: The brass, initially at a high temperature, will lose thermal energy as it cools down in the water.
• Heat Gain: Conversely, the water, starting off colder, absorbs thermal energy, increasing its temperature.
• Thermal Equilibrium: Over time, the brass and water reach a common temperature, termed thermal equilibrium, where no further heat transfer occurs between them.

The law of conservation of energy applies here, meaning that the total heat lost by the brass equals the heat gained by the water. This principle is crucial because it determines how we calculate the final temperature of the system.

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates the amount of heat required to change the temperature of one gram of the substance by one degree Celsius. The exercise illustrates how specific heat capacity plays a pivotal role in thermal processes.

Here's what you need to know about specific heat capacity:

• Brass: With a specific heat capacity of \(0.385 \text{ Jg}^{-1} \text{°C}^{-1}\), brass requires less energy to change its temperature compared to water.
• Water: Boasting a higher specific heat capacity of \(4.18 \text{ Jg}^{-1} \text{°C}^{-1}\), water can absorb more heat without a significant temperature increase.

In the context of the problem, the difference in specific heat capacities means water's temperature will change less compared to the brass when equal quantities of heat are exchanged. Recognizing these capacities helps us understand why the final equilibrium temperature is closer to the initial temperature of water.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. In this exercise, the insulated container acts as a calorimeter to ensure an accurate measurement of heat transfer.

Important points about calorimetry:

• Isolated System: The insulated container ensures that no heat escapes or enters the system, making heat exchange calculations more precise.
• Heat Exchange Equation: Using the equation \( Q = mcΔT \), where \( m \) is mass, \( c \) is specific heat capacity, and \( ΔT \) is the temperature change, helps determine the heat transferred between the brass and water.
• Solving for Temperature: By equating the heat lost by the brass to the heat gained by the water, we find a balance point, namely the final temperature \( T_{final} \).

Through proper calorimetry methods, accurate measurement of thermal interactions in the identified system allows the final temperature of the brass-water mixture to be calculated effectively.