Problem 60 - University Physics with Modern Physics

Question

A copper calorimeter can with mass 0.100 \(\mathrm{kg}\) contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Step-by-Step Solution

Answer

The final equilibrium temperature is approximately 26°C.

Step 1: Understand the Problem

We need to find the final temperature when heat is transferred between the copper calorimeter, water, ice, and lead as they reach thermal equilibrium. Use the concept of heat transfer and the principle of conservation of energy: the heat lost by hot lead will be equal to the heat gained by the ice, water, and calorimeter.

Step 2: List the Given Data and Constants

- Mass of calorimeter (copper), \( m_c = 0.100 \, \text{kg} \)- Mass of water, \( m_w = 0.160 \, \text{kg} \)- Mass of ice, \( m_i = 0.0180 \, \text{kg} \)- Mass of lead, \( m_l = 0.750 \, \text{kg} \)- Initial temperature of lead, \( T_l = 255^{\circ} \mathrm{C} \)- Specific heat capacity of copper, \( c_c = 385 \, \text{J/kg/}^{\circ} \mathrm{C} \)- Specific heat capacity of water, \( c_w = 4186 \, \text{J/kg/}^{\circ} \mathrm{C} \)- Specific heat capacity of ice, \( c_i = 2100 \, \text{J/kg/}^{\circ} \mathrm{C} \)- Specific heat capacity of lead, \( c_l = 128 \, \text{J/kg/}^{\circ} \mathrm{C} \)- Latent heat of fusion for ice, \( L_f = 334,000 \, \text{J/kg} \)

Step 3: Determine Heat Transfer for Each Component

For the lead cooling from \( 255^{\circ} \mathrm{C} \) to the final temperature \( T_f \), the heat lost by lead:\[ Q_{lead} = m_l \cdot c_l \cdot (T_i - T_f) \]For the copper calorimeter warming from \( 0^{\circ} \mathrm{C} \) (assumed initial equilibrium temperature) to \( T_f \):\[ Q_{cal} = m_c \cdot c_c \cdot (T_f - 0) \]For the water from \( 0^{\circ} \mathrm{C} \) to \( T_f \):\[ Q_{water} = m_w \cdot c_w \cdot (T_f - 0) \]For the ice melting and then warming to \( T_f \):\[ Q_{ice} = m_i \cdot L_f + m_i \cdot c_w \cdot (T_f - 0) \]

Step 4: Set Up the Energy Conservation Equation

According to the conservation of energy:\[ Q_{lead} = Q_{cal} + Q_{water} + Q_{ice} \]Substitute the expressions for \( Q_{lead}, Q_{cal}, Q_{water}, \text{and} \ Q_{ice} \):\[ m_l \cdot c_l \cdot (T_l - T_f) = m_c \cdot c_c \cdot T_f + m_w \cdot c_w \cdot T_f + m_i \cdot L_f + m_i \cdot c_w \cdot T_f \]

Step 5: Solve for the Final Temperature \( T_f \)

Rearrange to solve for \( T_f \):\[ m_l \cdot c_l \cdot (T_l - T_f) = (m_c \cdot c_c + m_w \cdot c_w + m_i \cdot c_w) \cdot T_f + m_i \cdot L_f \]Substitute the known values and solve for \( T_f \):- Left side with lead:\[ 0.750 \cdot 128 \cdot (255 - T_f) \]- Right side:\[ (0.100 \cdot 385 + 0.160 \cdot 4186 + 0.0180 \cdot 4186) \cdot T_f + 0.0180 \cdot 334,000 \]Simplify and solve the equation to find \( T_f \).

Step 6: Calculate the Final Temperature Numerically

Calculate each part:- \( Q_{lead} = 0.750 \times 128 \times (255 - T_f) \)- \( Q_{0.1\,\text{kg copper}} = 0.100 \times 385 \times T_f \)- \( Q_{0.16\,\text{kg water}} = 0.160 \times 4186 \times T_f \)- \( Q_{0.018\,\text{kg ice melts and heats}} = (0.018 \times 334000) + (0.018 \times 4186 \times T_f) \)Solving the complete equation gives \( T_f \approx 26.0^{\circ} \mathrm{C}\).

Step 7: Conclusion

Upon calculating, the final equilibrium temperature of the mixture when the lead is added to the calorimeter is found to be approximately \( 26.0^{\circ} \mathrm{C} \). This temperature ensures that all components have transferred heat until reaching thermal equilibrium with no external heat loss.

Key Concepts

Thermal Equilibrium

Thermal equilibrium is a fundamental concept in heat transfer. It occurs when two or more substances within a closed system exchange heat until they reach the same temperature. At this point, there is no net flow of thermal energy between them. This state ensures that all components in the system have achieved a balance in their heat energy. In our exercise, the copper calorimeter, water, ice, and lead achieve thermal equilibrium—their final temperature is the same.

This process illustrates the principle of heat transfer, where the heat lost by a hotter object, such as lead in our scenario, equals the heat gained by the colder components, like the ice, water, and calorimeter.

• It ensures no heat is gained or lost to the outside environment.
• All substances in contact exchange heat until their temperatures equalize.
• Once thermal equilibrium is reached, temperature variation stops, resulting in a stable system state.

Understanding thermal equilibrium is crucial for solving problems involving multiple substances interacting within a closed system, like the one described in this exercise.

Specific Heat Capacity

Specific heat capacity is an essential property of materials that measures how much heat energy a substance can absorb per unit mass per degree Celsius rise in temperature. It represents the resistance of a substance to temperature changes when heat is added or removed. In simpler terms, it's how much heat you need to warm up one kilogram of a material by one degree Celsius.

Each substance in the exercise—copper (calorimeter), water, ice, and lead—has its own specific heat capacity, influencing how each will respond to heat transfer. For instance:

• Copper has a lower specific heat capacity (\( 385 \; \text{J/kg/}^{\circ} \mathrm{C} \)), meaning it heats up and cools down more quickly than water.
• Water's high specific heat (\( 4186 \; \text{J/kg/}^{\circ} \mathrm{C} \)) means it requires more heat to change its temperature, which makes it effective for thermal storage.
• Lead, with its lower specific heat (\( 128 \; \text{J/kg/}^{\circ} \mathrm{C} \)), will rapidly alter temperature with heat exchange, being the source of heat in our exercise.

Using specific heat capacities allows us to calculate the heat each substance gives off or absorbs, aiding in the determination of the final equilibrium temperature.

Conservation of Energy

The principle of conservation of energy is pivotal in thermodynamics and states that energy cannot be created or destroyed; it can only be transformed or transferred. In the context of our exercise, this means that the total amount of heat energy in the system must remain constant. As such, the heat lost by the hot lead is equal to the sum of the heat gained by the copper calorimeter, the water, and the melting ice.

When the hot lead is introduced into the calorimeter:

• The lead cools down, transferring its thermal energy to bring all other components to thermal equilibrium.
• The calorimeter, water, and ice absorb this heat, warming up to the final equilibrium temperature.
• The mathematical formulation of this energy balance is shown in the equality set up between heat lost and heat gained.

This exchange ensures no heat energy escapes the system, aligning with the conservation principle. By applying this principle, we mathematically determine the final temperature through an energy balance equation, allowing us to solve for unknown temperatures and predict system behavior in thermal interactions.