Problem 39 - University Physics with Modern Physics

Question

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0\(^\circ\)C. A 0.250-kg block of iron at 85.0\(^\circ\)C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Step-by-Step Solution

Answer

The final temperature is approximately 26.5°C.

Step 1: Understand the Heat Exchange Concept

In this problem, the heat lost by the iron block will be equal to the heat gained by the pot and the water, because there is no heat loss to the surroundings.

Step 2: Write the Heat Exchange Equations

Use the formula for heat transfer, \[ Q = mc\Delta T \]where \( m \) is mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Write the equation for each component:1. Iron: \[ Q_{\text{iron}} = m_{\text{iron}} c_{\text{iron}} (T_{\text{final}} - T_{\text{initial}}) \]2. Water: \[ Q_{\text{water}} = m_{\text{water}} c_{\text{water}} (T_{\text{final}} - T_{\text{initial}}) \]3. Copper pot: \[ Q_{\text{copper}} = m_{\text{copper}} c_{\text{copper}} (T_{\text{final}} - T_{\text{initial}}) \]

Step 3: Set Up the Equality Equation

Because the heat lost by the iron is gained by the water and the copper pot, set up the equation as:\[ |Q_{\text{iron}}| = Q_{\text{water}} + Q_{\text{copper}} \]

Step 4: Substitute the Known Values

The specific heat capacities are:- Iron: \( c_{\text{iron}} = 450 \text{ J/(kg}\cdot^\circ\text{C)} \)- Water: \( c_{\text{water}} = 4186 \text{ J/(kg}\cdot^\circ\text{C)} \)- Copper: \( c_{\text{copper}} = 387 \text{ J/(kg}\cdot^\circ\text{C)} \)Substitute these values and the masses into the equations:\[ 0.250 \times 450 \times (T_{\text{final}} - 85) = 0.170 \times 4186 \times (T_{\text{final}} - 20) + 0.500 \times 387 \times (T_{\text{final}} - 20) \]

Step 5: Solve the Equation for Final Temperature

Expand and simplify the equation to solve for \( T_{\text{final}} \):1. \[ 112.5(T_{\text{final}} - 85) = 711.62(T_{\text{final}} - 20) + 193.5(T_{\text{final}} - 20) \]2. Simplify both sides and isolate \( T_{\text{final}} \).3. Solve to find \( T_{\text{final}} \approx 26.5^\circ\text{C} \)

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates how much energy is needed to change its temperature. It's a crucial factor when studying heat transfer. For any given material, the specific heat capacity (denoted as \( c \)) tells us how much heat (\( Q \)) is required to raise the temperature of a mass \( m \) by a certain change in temperature (\( \Delta T \)). The formula to calculate heat transfer using specific heat capacity is:

• \( Q = mc\Delta T \)

In our problem, we have three materials involved: iron, water, and copper (from the pot). Each of them has a different specific heat capacity:

• Iron: 450 J/(kg\cdot^\circ C)
• Water: 4186 J/(kg\cdot^\circ C)
• Copper: 387 J/(kg\cdot^\circ C)

This means that water requires more energy to change its temperature compared to copper and iron. Understanding these values allows us to determine how heat will flow between different materials when they are combined.

Thermal Equilibrium

Thermal equilibrium is the state reached when two or more bodies have exchanged heat to the point that there is no further net heat transfer between them. In other words, they've reached the same temperature. In this exercise, an iron block at a high temperature is placed into a cooler copper pot containing water. Initially, there is heat transfer:

• Iron releases heat as it cools down.
• The copper pot and water absorb this heat, warming up.

The system reaches thermal equilibrium when the final temperature is the same for all components. To find this equilibrium temperature, we balance the heat lost by the iron with the heat gained by the copper and water. This ensures no heat is lost to the surroundings, allowing us to determine that all the energy stays within the system until equilibrium is achieved.

Calorimetry

Calorimetry is the science of measuring the exchange of heat in physical and chemical processes. This concept is extremely valuable in experiments where temperature changes and heat transfer need to be analyzed. In the given exercise, we use calorimetry principles to determine the final temperature of the system comprising water, iron, and a copper pot. The setup involves:

• Calculating the heat lost by the iron as it cools down to thermal equilibrium.
• Calculating the heat gained by the copper pot and the water as they warm up.
• Setting the heat lost equal to the heat gained because no heat escapes the system.

By measuring initial temperatures, masses, and knowing specific heat capacities, calorimetry lets us solve for the unknown quantity, the final temperature. This problem showcases a typical calorimetry scenario where the final temperature is calculated by aligning the heat exchange in the materials involved.