Problem 50 - Chemistry: The Central Science in SI Units, Global Edition

Question

Two solid objects, A and B, are placed in boiling water and allowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing \(1000 \mathrm{~g}\) of water at \(10.0^{\circ} \mathrm{C}\). Object A increases the water temperature by \(3.50^{\circ} \mathrm{C} ; \mathrm{B}\) increases the water temperature by \(2.60{ }^{\circ} \mathrm{C}\). (a) Which object has the larger heat capacity? (b) What can you say about the specific heats of \(\mathrm{A}\) and \(\mathrm{B}\) ?

Step-by-Step Solution

Answer

The short answer to the question is: (a) Object A has the larger heat capacity, because it causes a greater temperature increase in the water it is placed in. (b) We cannot determine the specific heats of objects A and B from the given information, as we do not have their masses and initial temperatures.

Step 1: Find the heat transferred from the objects to the water

When each object is placed in the beakers, it transfers its heat to the water, which results in a temperature increase. We can use the equation: \[Q = mc\Delta T\] where \(Q\) is the heat transfer, \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Since the mass and the initial temperature of the water in the beakers are the same, we can compare the heat transfers (\(Q_A\), \(Q_B\)) by comparing their temperature increases (\(\Delta T_A\), \(\Delta T_B\)). **Step 2: Calculate the heat transfer from both objects**

Step 2: Calculate \(Q_A)\) and \(Q_B\)

From the given data, object A increases the water temperature by \(3.50^\circ \mathrm{C}\) and object B increases the water temperature by \(2.60{ }^{\circ} \mathrm{C}\). So, we can write \[\frac{Q_A}{Q_B} = \frac{\Delta T_A}{\Delta T_B}\] By plugging in the given values we get: \[\frac{Q_A}{Q_B} = \frac{3.50}{2.60}\] **Step 3: Compare the heat capacities**

Step 3: Determine which object has a larger heat capacity

Since the ratio \(\frac{Q_A}{Q_B} > 1\), this implies that object A loses more heat than object B, therefore it has a larger heat capacity. The answer of part (a) is: Object A has the larger heat capacity. **Step 4: Determine the specific heats of objects A and B**

Step 4: Determine if the specific heats of A and B can be compared

Specific heat is an intrinsic property of a material, meaning it is only dependent on the material itself and not the amount of the material. However, from the given data, we cannot determine the masses and initial temperatures of objects A and B, so it is impossible to compare their specific heats without more information. The answer to part (b) is: We cannot determine the specific heats of A and B from the given information.

Key Concepts

Understanding Specific Heat

Specific heat is a fundamental concept in thermodynamics. It tells us how much heat is needed to change the temperature of a certain amount of a substance. Specifically, it is defined as the amount of heat per unit mass required to raise the temperature by one degree Celsius. This property is intrinsic, meaning it doesn't change with the mass of the material.

When we talk about specific heat in everyday terms, we're essentially asking: "How much energy does it take to warm this material up?" Different materials might need different amounts of energy for the same temperature change. For example:

• Water has a high specific heat, meaning it takes a lot of energy to change its temperature.
• Metals, typically, have a lower specific heat, making them easier to heat up or cool down.

In the exercise, although we wanted to compare the specific heats of Objects A and B, the lack of details about their masses and initial temperatures hinders any direct comparison based solely on the heat they released to the water. Thus, specific heat remains constant for a substance but couldn't be determined here without extra information.

Exploring Heat Transfer

Heat transfer is the process of thermal energy passing from one object to another. It happens when there's a temperature difference between the two objects. In the context of our exercise, heat transfer occurs when Objects A and B are removed from boiling water and placed into room temperature water. The objects are hotter than the water in the beakers, so they release heat to the water until thermal equilibrium is reached.

The heat from the objects causes the water temperature to rise, which is an indication of energy being transferred. We use the formula \[ Q = mc\Delta T \] to calculate this energy, where:

• \( Q \) stands for the heat transferred,
• \( m \) represents the mass of the water,
• \( c \) is the specific heat capacity of water (usually known), and
• \( \Delta T \) is the change in the water's temperature.

In this scenario, the heat transfer is what allowed us to deduce that Object A had a larger heat capacity as it transferred more heat to the water, judging by the temperature change.

Determining Temperature Change

Temperature change in this context refers to the increase or decrease in the temperature of an object as it gains or loses heat. It is a key part of understanding how energy moves between objects. In the exercise, temperature change is observed when the objects (A and B) cause the water’s temperature to rise after submersion. This change provides insight into the amount of heat transferred.

The relation between heat transfer and temperature change is expressed by the formula:\[ Q = mc\Delta T \]Here, \( \Delta T \) is calculated as:

• \( \Delta T = \text{final temperature} - \text{initial temperature} \)

For example, in our exercise:

• Object A caused a temperature change of \( 3.50^{\circ} \text{C} \).
• Object B caused a temperature change of \( 2.60^{\circ} \text{C} \).

As a result, these numerical values directly helped in concluding which object transferred more heat based on how much they raised the water's temperature. Understanding temperature change is crucial for analyzing how different materials behave under various thermal conditions.