Problem 44 - College Physics

Question

\(\bullet\) 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 \(20.0^{\circ} \mathrm{C} . \mathrm{A} 0.250 \mathrm{kg}\) 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 \(27.2^{\circ} \mathrm{C}\).

Step 1: Identify known quantities

Begin by listing the known values for the materials involved in the problem. - Copper pot mass: \(m_c = 0.500\, \mathrm{kg}\)- Water mass: \(m_w = 0.170\, \mathrm{kg}\)- Iron block mass: \(m_i = 0.250\, \mathrm{kg}\)- Initial temperature of copper and water: \(T_{c,w} = 20.0^{\circ} \mathrm{C}\)- Initial temperature of iron: \(T_i = 85.0^{\circ} \mathrm{C}\)- Specific heat capacities: - Copper: \(c_c = 385\, \mathrm{J/kg^{\circ} C}\) - Water: \(c_w = 4186\, \mathrm{J/kg^{\circ} C}\) - Iron: \(c_i = 450\, \mathrm{J/kg^{\circ} C}\)

Step 2: Apply the conservation of energy principle

Since there is no heat loss to the surroundings, the heat gained by the copper pot and the water is equal to the heat lost by the iron block. Use the formula for heat transfer: \[ q = mc\Delta T \]where \( q \) is heat transfer, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is change in temperature.

Step 3: Set up the equation for heat exchange

Write the equation expressing conservation of energy:\[ m_c c_c (T_f - T_{c,w}) + m_w c_w (T_f - T_{c,w}) = m_i c_i (T_i - T_f) \]where \( T_f \) is the final temperature we need to find.

Step 4: Substitute known values into the equation

Substitute all known values:\[ 0.500 \times 385 \times (T_f - 20) + 0.170 \times 4186 \times (T_f - 20) = 0.250 \times 450 \times (85 - T_f) \]

Step 5: Simplify and solve the equation

First, simplify each term:- For the copper: \( 192.5(T_f - 20) \)- For the water: \( 711.62(T_f - 20) \)- For the iron: \( 112.5(85 - T_f) \)Combine the terms:\[ 192.5T_f - 3850 + 711.62T_f - 14232.4 = 112.5 \times 85 - 112.5T_f \]Combine like terms and solve for \( T_f \):\[ 904.12T_f - 18082.4 = 9562.5 - 112.5T_f \]\[ 1016.62T_f = 27644.9 \]\[ T_f \approx 27.2^{\circ} \mathrm{C} \]

Step 6: Evaluate the solution for consistency

Verify that the final temperature is reasonable. The final temperature \( T_f = 27.2^{\circ} \mathrm{C} \) is between the initial temperatures of the system, which reflects the expected mixture of temperatures considering no heat loss.

Key Concepts

Heat Transfer

Heat transfer is the process of thermal energy moving from a hotter object to a cooler one. In our exercise, the iron block at a higher temperature transfers heat to the copper pot and water, which are at a lower temperature. The energy moves as heat until thermal equilibrium is reached, meaning all components reach the same temperature.

The driving force behind heat transfer is the temperature difference between objects. Heat naturally flows from warm to cool until the temperature is uniform. Several modes of heat transfer exist, but in this problem, we consider conduction, a form of transfer where heat flows through direct contact between substances.

It's crucial to note that no external source affects the system in this problem, nor does any heat escape into the surroundings due to the assumption of an isolated system. This assumption simplifies calculations significantly and helps us understand practical applications such as cooking, where heat transfer often must be controlled to achieve desired results.

Specific Heat Capacity

Specific heat capacity ( ç) is a material's property defining how much energy it takes to change an object's temperature by one degree Celsius. Each material has a unique specific heat capacity, impacting how it exchanges heat. In simpler terms, it tells us how quickly or slowly a substance heats up or cools down. The higher an object's specific heat, the more heat it can absorb without changing temperature. In this problem: - Water has a specific heat capacity of 4186 J/kg°C, meaning it's slow to heat up or cool down. - Copper's specific heat capacity is 385 J/kg°C, showing it responds more readily to heat changes. - Iron, with a specific heat of 450 J/kg°C, heats and cools faster than water but slower than copper. Understanding specific heat capacity is crucial in predicting how different substances interact when they exchange heat, helping explain why the final temperature balances out as it does in the exercise. It's also widely applied in engineering and climate science to study heat absorption and release processes.

Conservation of Energy

The principle of the conservation of energy states that energy in a closed system remains constant—it can neither be created nor destroyed, only transferred or transformed. This exercise is a classic application, as all the heat lost by the hot iron block is entirely gained by the copper pot and water.The equation used in the exercise, \[m_c c_c (T_f - T_{c,w}) + m_w c_w (T_f - T_{c,w}) = m_i c_i (T_i - T_f)\],captures this principle by representing that the heat lost by iron equals the heat gained by copper and water. The left-hand side of the equation sums the absorbed heat, while the right-hand side gives the heat lost.This principle is fundamental in physics and engineering, ensuring calculations respect the physical world’s laws. It allows designers to predict system outcomes effectively, whether in thermal management systems or designing energy-efficient buildings.