Problem 83 - Chemistry and Chemical Reactivity

Question

A piece of lead with a mass of \(27.3 \mathrm{g}\) was heated to \(98.90^{\circ} \mathrm{C}\) and then dropped into \(15.0 \mathrm{g}\) of water at \(22.50^{\circ} \mathrm{C} .\) The final temperature was \(26.32^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of lead from these data.

Step-by-Step Solution

Answer

The specific heat capacity of lead is approximately \(0.121 \, \text{J/g}^{\circ}\text{C}\).

Step 1: Identify Known Variables

First, identify the known variables from the problem statement. The mass of lead, \( m_{\text{lead}} = 27.3 \, \text{g} \). The initial temperature of lead, \( T_{\text{initial, lead}} = 98.90^{\circ} \text{C} \). The final temperature of both substances, \( T_{\text{final}} = 26.32^{\circ} \text{C} \). The mass of water, \( m_{\text{water}} = 15.0 \, \text{g} \). The initial temperature of water, \( T_{\text{initial, water}} = 22.50^{\circ} \text{C} \). The specific heat capacity of water, \( c_{\text{water}} = 4.18 \, \text{J/g}^{\circ}\text{C} \).

Step 2: Set Up Heat Exchange Equation

Since no heat is lost to the surroundings, the heat lost by the lead will be equal to the heat gained by the water. This can be expressed as:\[m_{\text{lead}} \times c_{\text{lead}} \times (T_{\text{initial, lead}} - T_{\text{final}}) = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}})\]

Step 3: Solve Arithmetic for Water's Heat Gain

Calculate the heat gained by the water using its mass, specific heat capacity, and temperature change:\[q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}})\]Substitute the values:\[q_{\text{water}} = 15.0 \, \text{g} \times 4.18 \, \frac{\text{J}}{\text{g}^{\circ}\text{C}} \times (26.32^{\circ} \text{C} - 22.50^{\circ} \text{C})\]Calculate the expression:\[q_{\text{water}} = 15.0 \times 4.18 \times 3.82 = 239.43 \, \text{J}\]

Step 4: Calculate Specific Heat Capacity of Lead

Rearrange the heat exchange equation to find the specific heat capacity of lead:\[c_{\text{lead}} = \frac{m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}})}{m_{\text{lead}} \times (T_{\text{initial, lead}} - T_{\text{final}})}\]Substitute the known values:\[c_{\text{lead}} = \frac{239.43 \, \text{J}}{27.3 \, \text{g} \times (98.90^{\circ} \text{C} - 26.32^{\circ} \text{C})}\]Calculate the expression:\[c_{\text{lead}} = \frac{239.43}{27.3 \times 72.58} = \frac{239.43}{1981.014} = 0.121 \, \text{J/g}^{\circ}\text{C}\]

Key Concepts

Heat Exchange

Heat exchange is a fundamental concept in thermodynamics, where heat energy is transferred from one substance to another. This typically happens as the result of a temperature difference, where heat flows from the hotter object to the cooler one. When a hot object like a piece of lead is placed in cooler water, the lead loses heat while the water gains heat, until both reach a common temperature. This process continues until thermal equilibrium is achieved, meaning no more heat transfer occurs because the temperatures equalize.

In the given exercise, the lead piece at a higher temperature transfers heat to water. The calculation assumes all significant heat exchange happens between the lead and the water alone. Therefore, no heat is lost to the surroundings, making it a closed system. This simplicity allows us to set equations that equate the heat lost by the lead to the heat gained by the water, using their respective masses, specific heat capacities, and temperature changes.

Calorimetry

Calorimetry is the science of measuring heat exchange in physical and chemical processes. It involves precise calculations to determine the heat involved when substances absorb or release thermal energy. The calorimeter, the device used in these measurements, ideally isolates the experiment to ensure that no heat is lost to the environment and accurate readings can be achieved.

The exercise employs a basic calorimetry principle by using the known properties of water to discover the unknown specific heat capacity of lead. The method uses the formula:

• Heat lost by lead = Heat gained by water
• \[ m_{\text{lead}} \times c_{\text{lead}} \times (T_{\text{initial, lead}} - T_{\text{final}}) = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}}) \]

By substituting the known values for water’s mass, specific heat, and temperature change, we can solve for the unknown specific heat capacity of the lead.

Thermal Equilibrium

Thermal equilibrium occurs when two or more interacting substances reach the same temperature, leading to an equilibrium state with no further heat transfer occurring. This concept is crucial for many thermal calculations because it allows assumptions and simplifications in equations.

In the original exercise, thermal equilibrium is reached when the lead and water reach a final common temperature of \(26.32^{\circ} \text{C}\). At this point, the heat lost by the lead equals the heat gained by the water, ensuring no more heat transfer occurs. This equilibrium point is key for calculating heat exchange, as it signifies the process' end, where all transactional heat has been balanced between the involved substances.