Question

A 192 -g piece of copper is heated to \(100.0^{\circ} \mathrm{C}\) in a boiling water bath and then dropped into a beaker containing \(\left.751 \mathrm{g} \text { of water (density }=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\) at \(4.0^{\circ} \mathrm{C} .\) What is the final temperature of the copper and water after thermal equilibrium is reached? (The specific heat capacity of copper is \(0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K})\).

Short answer

The final temperature is approximately 6.21°C.

Step-by-step solution

Identify the Known Values

We have two substances: copper and water. Their initial conditions are as follows:- Mass of copper, \(m_c = 192 \text{ g}\), initial temperature \(T_{c,i} = 100.0^{\circ}\text{C}\).- Mass of water, \(m_w = 751 \text{ g}\), initial temperature \(T_{w,i} = 4.0^{\circ}\text{C}\).- Specific heat capacity of copper \(C_c = 0.385 \text{ J/g}\cdot\text{K}\).- Specific heat capacity of water \(C_w = 4.18 \text{ J/g}\cdot\text{K}\).

Set Up the Heat Exchange Equation

At thermal equilibrium, the heat lost by copper will equal the heat gained by the water:\[-(m_c \cdot C_c \cdot (T_f - T_{c,i})) = m_w \cdot C_w \cdot (T_f - T_{w,i})\] where \(T_f\) is the final temperature, which is the same for copper and water when equilibrium is reached.

Plug in the Known Values to the Equation

Substituting in the known values:\[-(192 \cdot 0.385 \cdot (T_f - 100)) = 751 \cdot 4.18 \cdot (T_f - 4)\] Simplify the expression to:\[-(73.92 \cdot (T_f - 100)) = 3142.18 \cdot (T_f - 4)\].

Solve for the Final Temperature \(T_f\)

Distribute and simplify both sides:\[-73.92T_f + 7392 = 3142.18T_f - 12568.72\].Rearrange to group \(T_f\) terms:\[73.92T_f + 3142.18T_f = 7392 + 12568.72\].This simplifies to:\[3216.1T_f = 19960.72\].Divide to solve for \(T_f\):\[T_f = \frac{19960.72}{3216.1} \approx 6.21^{\circ}\text{C}\].

Key concepts

Heat Transfer

Heat transfer is a fundamental concept when it comes to understanding how different materials interact thermally. In this scenario, when we place the hot copper in cold water, an energy exchange occurs between the two. This energy movement is driven by the principle that heat flows from the warmer object (copper) to the cooler one (water) until they both reach the same temperature, known as thermal equilibrium.
This process in our example can be explained through:

• Conduction: The mode of heat transfer here involves direct contact between copper and water, allowing heat energy from copper atoms to transfer to water molecules.
• Conservation of Energy: The total energy in the system remains constant, meaning the heat lost by copper must be gained by water.

Understanding this basic heat transfer principle helps us predict how the final temperature (when an equilibrium is reached) can be calculated.

Specific Heat Capacity

The specific heat capacity is a measure of how much energy is needed to raise the temperature of a material. It plays a crucial role in determining how substances react to thermal energy exchanges.
In the context of our problem, consider:

• Copper's specific heat capacity is 0.385 J/g·K. This relatively low value indicates that copper doesn't require much energy to change its temperature.
• In contrast, water has a higher specific heat capacity of 4.18 J/g·K. This suggests that water needs a larger amount of energy to experience the same temperature change as copper.

These specific heat capacities are essential in the equation we use to set up the system's heat exchange, establishing how much energy is transferred during the temperature changes until thermal equilibrium is reached.

Final Temperature Calculation

To find the final temperature when copper and water reach thermal equilibrium, we need to use a heat exchange equation that factors in their specific heat capacities and initial conditions.
Using the equation \[-(m_c \cdot C_c \cdot (T_f - T_{c,i})) = m_w \cdot C_w \cdot (T_f - T_{w,i})\]we compute the heat lost by copper and the heat gained by water being equal, ensuring the balance of energy.
Here's how it breaks down:

• On substitution, the heat balance equation becomes \[-(192 \cdot 0.385 \cdot (T_f - 100)) = 751 \cdot 4.18 \cdot (T_f - 4)\]
• Continue to distribute and simplify to pull together unknowns related to the final temperature \(T_f\).
• Solving the simplified equation \[3216.1T_f = 19960.72\], by dividing both sides, yields the final temperature of \(T_f = 6.21^{\circ} \text{C}\).

Understanding the calculation step-by-step shows that each parameter—mass, specific heat, and initial temperatures—affects the final outcome, allowing the substances to reach a shared temperature.