Question

A piece of gold \(\left(10.0 \mathrm{g}, C_{\mathrm{Au}}=0.129 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\) is heated to \(100.0^{\circ} \mathrm{C} .\) A piece of copper (also \(10.0 \mathrm{g}\), \(\left.C_{\mathrm{Cu}}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\) is chilled in an ice bath to \(0^{\circ} \mathrm{C}\) Both pieces of metal are placed in a beaker containing \(150 . \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C} .\) Will the temperature of the water be greater than or less than \(20^{\circ} \mathrm{C}\) when thermal equilibrium is reached? Calculate the final temperature.

Short answer

The final temperature is about 20.1°C, so the water temperature will increase slightly.

Step-by-step solution

Identify the Heat Transfer Equation

When two objects at different temperatures come into contact, heat flows from the hotter object to the cooler one until thermal equilibrium is reached. We can use the formula for heat transfer: \[ q = mc\Delta T \]where \( q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Set Up Heat Transfer Equations for Each Substance

We have three components: gold, copper, and water. - For gold cooling down: \[ q_{\text{Au}} = m_{\text{Au}} c_{\text{Au}} (T_f - T_{\text{Au initial}}) \]- For copper heating up:\[ q_{\text{Cu}} = m_{\text{Cu}} c_{\text{Cu}} (T_f - T_{\text{Cu initial}}) \]- For water:\[ q_{\text{H}_2\text{O}} = m_{\text{H}_2\text{O}} c_{\text{H}_2\text{O}} (T_f - T_{\text{H}_2\text{O initial}}) \]

Apply the Conservation of Energy Principle

The principle of conservation of energy states that the total heat gained by the gold and copper is equal to the heat lost by the water. Thus, \[ q_{\text{Au}} + q_{\text{Cu}} = -q_{\text{H}_2\text{O}} \]The left side represents the heat gains by the metals, and the right side represents the heat loss by the water.

Substitute Known Values

Substitute the known masses, specific heat capacities, and initial temperatures into the heat transfer equations:- Gold: \( m_{\text{Au}} = 10.0 \) g, \( c_{\text{Au}} = 0.129 \) J/g·K, \( T_{\text{Au initial}} = 100.0 \) °C- Copper: \( m_{\text{Cu}} = 10.0 \) g, \( c_{\text{Cu}} = 0.385 \) J/g·K, \( T_{\text{Cu initial}} = 0 \) °C- Water: \( m_{\text{H}_2\text{O}} = 150.0 \) g, \( c_{\text{H}_2\text{O}} = 4.18 \) J/g·K, \( T_{\text{H}_2\text{O initial}} = 20.0 \) °C.

Solve the Equation for Final Temperature

Plug the values into the conservation equation:\[10.0 \times 0.129 \times (T_f - 100.0) + 10.0 \times 0.385 \times (T_f - 0) = -150.0 \times 4.18 \times (T_f - 20.0)\]Solve for \( T_f \) to find the final equilibrium temperature for the system.

Calculate the Final Temperature (T_f)

Simplify and solve the equation:- Expand and combine terms:\[ 1.29(T_f - 100) + 3.85T_f = -627(T_f - 20) \]- Distribute: \[ 1.29T_f - 129 + 3.85T_f = -627T_f + 12540 \]- Combine like terms:\[ 5.14T_f - 129 = -627T_f + 12540\]- Add \(627T_f\) to both sides and solve for \(T_f\):\[632.14T_f = 12669 \] \[T_f \approx 20.1\text{°C}\]

Interpret the Result

The calculation gives a final temperature \(T_f = 20.1\text{°C}\). Because this value is slightly greater than the initial temperature of water, the temperature of the water will increase, albeit by a small amount.

Key concepts

Specific Heat Capacity

Specific heat capacity is a concept you encounter when dealing with heat transfer. It's a measure of how much heat energy is needed to raise the temperature of a given mass of a substance by one degree Celsius (or one Kelvin). This property is crucial because it determines how substances absorb and release heat. For example, in the problem, gold and copper have different specific heat capacities:

• Gold: 0.129 J/g·K
• Copper: 0.385 J/g·K

These numbers tell you that copper needs more energy to change its temperature compared to gold. This difference plays a significant role in calculating how each metal affects the temperature of the water when they are mixed.

Thermal Equilibrium

Thermal equilibrium is the state you reach when two objects in thermal contact no longer exchange heat because they've reached the same temperature. Imagine placing the heated gold and chilled copper into the water. These substances will continue to exchange heat with each other and the water until they all stabilize at a uniform temperature, where thermal equilibrium is achieved.

This process, known as reaching thermal equilibrium, highlights the flow of heat from the hotter object (gold) to the cooler one (copper and water) until balance is struck, and no further temperature change occurs.

Conservation of Energy

In the context of heat transfer, the conservation of energy principle states that the total energy in a closed system remains constant. In simpler terms, the energy lost by one part of the system must equal the energy gained by another.

Applying this to our example, as the gold piece loses heat, the copper and water gain that same amount of heat, resulting in no net change in the total energy of the system. This idea governs the setup of the equation where the heat lost by the gold must precisely equal the heat gained by the copper and water. That mathematical equation becomes the tool for calculating the final balanced temperature of the whole system.

Temperature Change Calculations

Temperature change calculations involve using the heat transfer formula to assess how much the temperature of an object will change when subjected to gaining or losing heat. The formula used is: \[ q = mc\Delta T \] where \( q \) is the heat energy transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

In our problem, this formula helps determine how much the temperatures of gold, copper, and water will change until they reach equilibrium. By substituting each substance's mass, specific heat capacity, and temperature change into this formula, you can deduce how heat is exchanged among the different components, eventually calculating the final temperature after all heat transfer has occurred.