Problem 44 - College Physics

Question

A copper pot with a mass of \(0.500 \mathrm{~kg}\) contains \(0.170 \mathrm{~kg}\) of water, and both are at a temperature of \(20.0^{\circ} \mathrm{C}\). A \(0.250 \mathrm{~kg}\) block of iron at \(85.0^{\circ} \mathrm{C}\) 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 of the system is approximately 20.56°C.

Step 1: Identify Components and Their Specific Heat Capacities

The problem involves three components: copper pot, water, and the iron block. Their specific heat capacities are:- Copper: \( c_{cu} = 385 \, \text{J/kg°C} \)- Water: \( c_w = 4186 \, \text{J/kg°C} \)- Iron: \( c_{fe} = 450 \, \text{J/kg°C} \)These values will be used to calculate heat exchange.

Step 2: Set Up Heat Exchange Equation

Since no heat is lost to the surroundings, the heat lost by the iron will be gained by the copper pot and water. Thus, the heat exchange equation becomes:\[m_{fe}c_{fe}(T_i - T_f) = m_{cu}c_{cu}(T_f - T_i) + m_wc_w(T_f - T_i)\]where \(m\) is mass, \(c\) is specific heat, \(T_i\) and \(T_f\) are initial and final temperatures.

Step 3: Substitute Values into the Equation

Substitute the following values into the heat exchange equation:- For iron: \( m_{fe} = 0.250 \, \text{kg} \), \( T_i = 85.0 \, °C \)- For copper: \( m_{cu} = 0.500 \, \text{kg} \), \( T_i = 20.0 \, °C \)- For water: \( m_w = 0.170 \, \text{kg} \), \( T_i = 20.0 \, °C \)Thus the equation becomes:\[0.250 \times 450 \times (85 - T_f) = 0.500 \times 385 \times (T_f - 20) + 0.170 \times 4186 \times (T_f - 20)\]

Step 4: Solve for Final Temperature \(T_f\)

Simplify and solve the equation:\[112.5 (85 - T_f) = 192.5 (T_f - 20) + 711.62 (T_f - 20)\]Calculate:\[112.5 \times 85 - 112.5T_f = 192.5T_f - 192.5 \times 20 + 711.62T_f - 711.62 \times 20\]Combine terms and solve for \(T_f\).

Step 5: Calculate and Conclude Final Temperature

After solving the equation, we find that:\[112.5 \times 85 - 112.5T_f = 192.5 T_f + 711.62 T_f - 9037.6 \]which implies:\[9562.5 = 904.12 T_f - 9037.6 \]\[ 18600.5 = 904.12 T_f \]\[T_f \approx 20.56 \, °C\]

Step 6: Final step: Verification

Ensure calculations and assumptions are correct. Since temperatures fall in reasonable range, calculate and check again if needed.

Key Concepts

Specific Heat Capacity

In physics, specific heat capacity is a critical concept that represents the amount of heat required to change the temperature of 1 kilogram of a substance by 1 degree Celsius. Each material has a unique specific heat capacity, which is a measure of how efficiently it absorbs heat.

• Copper has a specific heat capacity of 385 J/kg°C, which means it requires 385 Joules to raise 1 kg of copper by 1°C.
• Water, known for its high specific heat capacity of 4186 J/kg°C, can absorb a lot of heat without a significant change in temperature. This property makes water an excellent medium for heat transfer.
• Iron, with a specific heat capacity of 450 J/kg°C, can also store and transfer heat, although not as efficiently as water.

Understanding these values helps in comprehending how much heat energy each material will absorb or release during a temperature change.

Thermal Equilibrium

Thermal equilibrium is achieved when two or more objects reach the same temperature and no longer exchange heat. In this exercise, the copper pot, water, and iron block exchange heat until they reach a uniform temperature.
Initially, the iron block is hotter than both water and the copper pot. As heat transfers from the iron to the other components, the temperatures adjust. Eventually, all elements settle at a common final temperature, called thermal equilibrium.
Achieving thermal equilibrium means that the total heat lost by the hot object equals the total heat gained by the cooler objects. This concept is essential in solving problems involving mixed substances and changing temperatures.

Calorimetry

Calorimetry is the science of measuring heat transfer within a system. It involves calculating the heat exchanged between different substances to determine changes in energy and temperature.
In this problem, calorimetry is used to estimate the final temperature when objects of different initial temperatures come into contact. By applying the principle of energy conservation, calorimetry allows us to understand how energy is distributed.
The essential calorimetry equation here accounts for the heat lost by the iron block, which is equal to the heat gained by the copper pot and water. This balance allows us to determine the final temperature of the system.

Energy Conservation

Energy conservation is a fundamental principle stating that energy cannot be created or destroyed in an isolated system. Instead, it is transferred from one form to another.
In our context, the principle of energy conservation is applied to the heat exchange between materials. The energy lost by the iron block as it cools is completely transferred to the copper pot and water, raising their temperatures.
This principle simplifies solving for the final temperature by ensuring that the total energy remains constant. Understanding energy conservation helps in interpreting how heat energy is redistributed in different parts of a problem.