Problem 94 - Chemistry and Chemical Reactivity

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 751 g of water (density = \(1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) at \(4.0^{\circ} \mathrm{C}\). What was the final temperature of the copper and water after thermal equilibrium was reached? \(\left(C_{C u}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\right).\)

Step-by-Step Solution

Answer

The final temperature is approximately 6.21°C.

Step 1: Identify the known quantities

The known quantities are:- Mass of copper, \(m_{\text{Cu}} = 192 \text{ g}\)- Initial temperature of copper, \(T_{\text{Cu, initial}} = 100.0^{\circ} \text{C}\)- Specific heat capacity of copper, \(C_{\text{Cu}} = 0.385 \text{ J/g} \cdot \text{K}\)- Mass of water, \(m_{\text{water}} = 751 \text{ g}\)- Initial temperature of water, \(T_{\text{water, initial}} = 4.0^{\circ} \text{C}\)- Specific heat capacity of water, \(C_{\text{water}} = 4.18 \text{ J/g} \cdot \text{K} \) (This is a standard value)- The density of water, \(\text{density} = 1.00 \text{ g/cm}^3\) is not used in the calculations as volume doesn't change.

Step 2: Set up the equation for heat exchange

When the copper and water reach thermal equilibrium, the heat lost by copper equals the heat gained by the water. This can be expressed as:\[ m_{\text{Cu}} \cdot C_{\text{Cu}} \cdot (T_{\text{final}} - T_{\text{Cu, initial}}) = - m_{\text{water}} \cdot C_{\text{water}} \cdot (T_{\text{final}} - T_{\text{water, initial}}) \]

Step 3: Substitute known values into the equation

Substitute the given values into the heat exchange equation:\[ 192 \cdot 0.385 \cdot (T_{\text{final}} - 100.0) = - 751 \cdot 4.18 \cdot (T_{\text{final}} - 4.0) \]

Step 4: Simplify and solve for final temperature

First, simplify both sides:\[ 73.92(T_{\text{final}} - 100) = -3140.18(T_{\text{final}} - 4) \]Expand both sides:\[ 73.92T_{\text{final}} - 7392 = -3140.18T_{\text{final}} + 12560.72 \]Now, add \(3140.18T_{\text{final}}\) to both sides:\[ 3214.10T_{\text{final}} = 12560.72 + 7392 \]Combine terms on the right:\[ 3214.10T_{\text{final}} = 19952.72 \]Divide by 3214.10 to find \(T_{\text{final}}\):\[ T_{\text{final}} = \frac{19952.72}{3214.10} \approx 6.21^{\circ} C \]

Step 5: Conclude with the final answer

The final temperature of the copper and water, after reaching thermal equilibrium, is approximately \(6.21^{\circ} \text{C}\).

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from a hotter object to a cooler one. In this exercise, we observed heat transfer between a piece of copper and water. When these materials are combined, the copper loses heat, while the water gains it. This exchange continues until both reach the same temperature, known as thermal equilibrium, where their temperatures no longer change.

This phenomenon is governed by the principle that heat will naturally flow from high to low temperature. Understanding this is crucial in various scientific fields, particularly in solving problems around thermal equilibrium, where the energy lost by a hot object equals the energy gained by a cold object. This balance allows us to predict the final temperatures after mixing.

Specific Heat Capacity

Specific heat capacity is a measure of how much energy is required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). In our example, copper has a specific heat capacity of 0.385 J/g·K, while water has 4.18 J/g·K. This means that water requires more energy to change its temperature compared to copper.

The specific heat capacity plays a vital role in calculating thermal interactions. It determines how a substance will react to thermal energy gain or loss. In the context of the problem, knowing these values allows us to set up an equation to find the final equilibrium temperature.

Temperature Change

Temperature change is the difference in temperature that a substance undergoes when it gains or loses heat. In the heat exchange equation, this is represented by \(T_{\text{final}} - T_{\text{initial}}\). For the copper, the initial temperature was higher, and for the water, it was lower. Their respective changes depend on their specific heat capacities and masses.

This concept is quantifiable and crucial for solving calorimetry problems. By understanding how temperature changes with added or subtracted energy, we can determine the new states of substances involved in the exchange.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It relies on understanding the principles of heat transfer, specific heat capacity, and temperature change. In this exercise, a calorimetric approach was used to determine the final temperature when copper is mixed with water.

To solve calorimetry problems, one sets up equations based on the conservation of energy. The heat lost by the warmer object is equal to the heat gained by the cooler one. This balance allows us to find unknown values, such as the final equilibrium temperature, as we did with the copper and water.