Problem 114 - Chemistry and Chemical Reactivity

Question

A A piece of gold \(\left(10.0 \mathrm{g}, C_{\mathrm{Au}}=0.129 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\) is heated to \(100.0^{\circ} \mathrm{C} .\) A piece of copper (also \(10.0 \mathrm{g}\) \(\left.C_{\alpha_{i}}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\) is chilled in an ice bath to \(0^{\circ} \mathrm{C} .\) Both pieces of metal are placed in a beaker containing \(150 . \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C} .\) Will the temperature of the water be greater than or less than \(20^{\circ} \mathrm{C}\) when thermal equilibrium is reached? Calculate the final temperature.

Step-by-Step Solution

Answer

The water temperature will be less than \(20^{\circ}C\) at equilibrium, final temp is \(-7.68^{\circ}C\).

Step 1: Determine heat lost by gold

The formula to calculate heat lost or gained, \( q \), is given by \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is change in temperature. For gold, \( q_{\text{Au}} = (10.0 \, \text{g}) \times (0.129 \, \text{J/g} \cdot \text{K}) \times (T_f - 373.15 \, \text{K}) \), where \( T_f \) is the final temperature.

Step 2: Determine heat gained by copper

Similarly, calculate the heat gained by copper. For copper, \( q_{\text{Cu}} = (10.0 \, \text{g}) \times (0.385 \, \text{J/g} \cdot \text{K}) \times (T_f - 273.15 \, \text{K}) \).

Step 3: Determine heat gained by water

The water also gains or loses heat. For water, \( q_{\text{H}_2\text{O}} = (150.0 \, \text{g}) \times (4.186 \, \text{J/g} \cdot \text{K}) \times (T_f - 293.15 \, \text{K}) \).

Step 4: Apply conservation of energy

According to the principle of conservation of energy, the sum of heat changes in the system is zero: \( q_{\text{Au}} + q_{\text{Cu}} + q_{\text{H}_2\text{O}} = 0 \). Substituting in the expressions from previous steps gives:\[(10.0)(0.129)(T_f - 373.15) + (10.0)(0.385)(T_f - 273.15) + (150)(4.186)(T_f - 293.15) = 0.\]

Step 5: Solve the equation for \( T_f \)

Expand the equation to combine terms and solve for \( T_f \):\[(1.29T_f - 481.0465) + (3.85T_f - 1050.0775) + (627.9T_f - 1840.65) = 0.\]Combine like terms:\[632.97T_f - 3371.774 = 0.\]Rearrange to solve:\[632.97T_f = 3371.774,\]\[T_f = \frac{3371.774}{632.97} = 5.32 \, \text{K}.\]The calculation shows the final temperature in Kelvin, which is 265.47 K equivalent to \(-7.68^{\circ}C\).

Key Concepts

Heat Transfer

Whenever two objects at different temperatures come into contact, heat transfer occurs. This is a process where thermal energy moves from an object with higher temperature to one with lower temperature. In this exercise, gold is initially at a high temperature of 100°C, while copper begins at 0°C. Both are placed in water at 20°C. Here, the flow of thermal energy will happen from the warmer substances to the cooler ones until they all reach the same temperature, a state known as thermal equilibrium.

• The gold, being the hottest, will lose heat.
• The copper and water, at lower temperatures, will gain heat.

This pool of energy sharing helps all substances involved to equalize in temperature, demonstrating the system reaching thermal equilibrium.

Specific Heat Capacity

Specific heat capacity is a property of a material that describes how much heat energy is required to change its temperature by one degree Celsius (or Kelvin). This is a crucial factor in predicting how substances will respond to heat transfer.

• Gold, in the present example, has a specific heat capacity of 0.129 J/g·K.
• Copper shows a comparatively higher specific heat capacity of 0.385 J/g·K.
• Water's specific heat capacity is 4.186 J/g·K, substantially higher than either metal.

Due to water's high specific heat capacity, it can absorb a large amount of heat without a significant change in temperature, playing a major role in stabilizing the temperature during the heat exchange process.

Conservation of Energy

The principle of conservation of energy states that energy in a closed system cannot be created or destroyed; it can only be transferred or transformed. This exercise uses energy conservation to predict the final temperature of the system. When the gold, copper, and water exchange heat, the total heat change in the system remains zero.

Mathematically, this is captured in the equation:

• \[ (q_{\text{Au}}) + (q_{\text{Cu}}) + (q_{\text{H}_2\text{O}}) = 0 \]

Here, each \( q \) term represents the heat lost or gained by the respective substance. The solution involves solving this equation, determining that the final equilibrium temperature of the system is lower than the initial temperature of the water, at \(-7.68^\circ C\) or 265.47 K. Each involved material shares energy until none have any heat energy to gain or lose, demonstrating conservation of energy principles in action.