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.

Short answer

The specific heat capacity of lead is approximately 0.121 J/g°C.

Step-by-step solution

Understand the Principle

The heat lost by the lead as it cools is gained by the water. Using the concept of conservation of energy, we can set up the equation: \[ q_{lead} = q_{water} \] where \(q\) is the heat exchanged.

Calculate Heat Gained by Water

Use the formula for heat transfer: \[ q = m \, c \, \Delta T \]where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.For water: - Mass \(m = 15.0 \, \text{g}\)- \(\Delta T = 26.32^{\circ} \mathrm{C} - 22.50^{\circ} \mathrm{C} = 3.82^{\circ} \mathrm{C}\)- \(c_{water} = 4.18 \, \mathrm{J/g^{\circ}C}\)Calculate the heat gained by water: \[ q_{water} = 15.0 \, \mathrm{g} \times 4.18 \, \mathrm{J/g^{\circ}C} \times 3.82^{\circ} \mathrm{C} \approx 239.43 \, \mathrm{J} \]

Set Up Equation for Lead

The heat lost by the lead is equal to the heat gained by water:\[ q_{lead} = m_{lead} \, c_{lead} \, \Delta T_{lead} \]Given:- \(m_{lead} = 27.3 \, \mathrm{g}\)- Initial Temperature \(T_{initial} = 98.90^{\circ} \mathrm{C}\)- Final Temperature \(T_{final} = 26.32^{\circ} \mathrm{C}\)- \(\Delta T_{lead} = 98.90^{\circ} \mathrm{C} - 26.32^{\circ} \mathrm{C} = 72.58^{\circ} \mathrm{C}\)Substituting into the equation:\[ 239.43 = 27.3 \, c_{lead} \, \times 72.58 \]

Solve for Specific Heat Capacity of Lead

Rearrange the equation to solve for \(c_{lead}\): \[ c_{lead} = \frac{239.43}{27.3 \times 72.58} \]Calculate:\[ c_{lead} \approx \frac{239.43}{1982.034} \approx 0.121 \, \mathrm{J/g^{\circ}C} \]

Conclusion

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

Key concepts

Conservation of Energy

In physics, the principle of conservation of energy states that energy cannot be created or destroyed, but can only change from one form to another. This is a foundational concept in understanding how energy transfers between systems. For instance, when a hot object like lead is placed in contact with cooler water, heat energy from the lead is transferred to the water until they reach a thermal equilibrium.

The heat lost by one object is equal to the heat gained by another in a closed system. In our specific problem, the heat lost by the hot lead must entirely match the heat gained by the cooler water. This balance of energy illustrates the conservation of energy principle at work. By setting the heat lost equal to the heat gained, we can solve for unknown quantities such as the specific heat capacity of materials involved.

Heat Transfer Formula

Heat transfer is the process by which heat energy moves from one object to another, and it can be calculated using a simple formula. The formula for heat transfer is \( q = m \, c \, \Delta T \), where:

• \( q \) is the amount of heat transferred, measured in joules \( \mathrm{J} \)
• \( m \) is the mass of the substance, in grams \( \mathrm{g} \)
• \( c \) is the specific heat capacity \( \mathrm{J/g^{\circ}C} \)
• \( \Delta T \) is the temperature change experienced by the substance \( \mathrm{^{\circ}C} \)

This equation allows us to calculate how much heat is exchanged between two substances. In our problem, we used this formula to determine the heat gained by the water. By knowing the mass of the water, its specific heat capacity, and the change in temperature, we calculated the specific amount of energy transferred.

Temperature Change

Temperature change \( \Delta T \) is a crucial factor in calculating heat transfer, as it represents the difference in temperature that a substance undergoes during heating or cooling. In calorimetry, accurately measuring this change is essential for determining energy transactions within the system.

For the lead and water calorimetry problem, we evaluate the change in temperature for both substances. The lead initially at \(98.90^{\circ} \mathrm{C}\) cooled down to meet the final equilibrium temperature of \(26.32^{\circ} \mathrm{C}\). Similarly, the water heated up, starting from \(22.50^{\circ} \mathrm{C}\).

The temperature changes are \(72.58^{\circ} \mathrm{C}\) for lead and \(3.82^{\circ} \mathrm{C}\) for water. These changes are key to applying the heat transfer formula. Proper calculation and interpretation of \( \Delta T \) help in understanding how heat flows between objects.

Calorimetry

Calorimetry involves measuring the amount of heat absorbed or released during a chemical or physical process. In our exercise, we conducted a basic calorimetry experiment to find the specific heat capacity of lead by observing heat exchanges between it and water.

The calorimeter, in its simplest form, is a container where the reaction or transfer occurs. By ensuring no heat escapes to the surroundings, we assume an isolated system where all heat loss or gain is confined within the materials involved.

Performing calorimetry requires careful consideration of initial and final temperatures, masses, and specific heat capacities. With these, the total heat exchange can be computed to yield insights into unknown thermal properties, like finding the specific heat capacity of lead which was calculated to be approximately \(0.121 \, \mathrm{J/g^{\circ}C}\). Calorimetry thus provides a practical method for exploring thermodynamic principles in the laboratory.