Problem 15 - Chemistry and Chemical Reactivity

Question

A 13.8 -g piece of zinc was heated to \(98.8^{\circ} \mathrm{C}\) in boiling water and then dropped into a beaker containing \(45.0 \mathrm{g}\) of water at \(25.0^{\circ} \mathrm{C} .\) When the water and metal came to thermal equilibrium, the temperature was \(27.1^{\circ} \mathrm{C} .\) What is the specific heat capacity of zinc?

Step-by-Step Solution

Answer

The specific heat capacity of zinc is approximately 0.398 J/g°C.

Step 1: Identify the known values

First, collect the known values from the problem. The mass of zinc is given as 13.8 g, the initial temperature of zinc is 98.8°C, and the final equilibrium temperature is 27.1°C. The mass of the water is 45.0 g, its initial temperature is 25.0°C, and its final temperature is also 27.1°C.

Step 2: Calculate the heat absorbed by water

Use the formula for heat absorbed: \[ q = m \cdot c \cdot \Delta T \]where \( q \) is the heat absorbed, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.Given: - Mass of water, \( m = 45.0 \text{ g} \)- Specific heat capacity of water, \( c = 4.18 \text{ J/g°C} \)- Temperature change, \( \Delta T = 27.1 - 25.0 = 2.1 \text{ °C} \)Calculate the heat absorbed by the water: \[ q = 45.0 \times 4.18 \times 2.1 \approx 396.81 \text{ J} \]

Step 3: Apply the principle of conservation of energy

Assume there is no heat loss to the surroundings, the heat lost by the zinc is equal to the heat gained by the water. Thus, the heat lost by zinc is also 396.81 J.

Step 4: Calculate the specific heat capacity of zinc

Using the formula for heat lost: \[ q = m \cdot c \cdot \Delta T \]Rearrange it for \( c \) (the specific heat capacity of zinc): \[ c = \frac{q}{m \cdot \Delta T} \]Given:- Heat lost by zinc, \( q = 396.81 \text{ J} \)- Mass of zinc, \( m = 13.8 \text{ g} \)- Temperature change for zinc, \( \Delta T = 98.8 - 27.1 = 71.7 \text{ °C} \)Calculate \( c \):\[ c = \frac{396.81}{13.8 \times 71.7} \approx 0.398 \text{ J/g°C} \]

Step 5: Conclusion

The specific heat capacity of zinc is computed to be approximately \( 0.398 \text{ J/g°C} \).

Key Concepts

Thermal Equilibrium

Thermal equilibrium is a state where two or more substances reach a common temperature. This occurs when there is no net heat transfer between them. In our example, the zinc and water reach thermal equilibrium at \(27.1^{\circ}\)C. This means the zinc has cooled down, while the water has warmed up, until their temperatures equalize.

Understanding thermal equilibrium involves recognizing that heat moves from the hotter object (zinc) to the cooler one (water) until their temperatures match.

• In the example problem, the zinc starts at \(98.8^{\circ}\)C and the water at \(25.0^{\circ}\)C.
• After the transfer, both reach \(27.1^{\circ}\)C.

Reaching thermal equilibrium doesn't result in any further temperature change unless additional heat transfer occurs from other sources.

Conservation of Energy

The principle of conservation of energy states that in a closed system, energy cannot be created or destroyed, only transformed. In the context of heat transfer, this principle tells us that any heat lost by a substance must be gained by another, unless lost to the surroundings.

In our exercise, the closed system includes the piece of zinc and the water.

• The zinc loses heat, cooling as it transfers thermal energy to the water.
• The water absorbs this heat, increasing in temperature.

Despite the transfer, the total amount of energy remains constant, aligning with the conservation of energy.

This concept underpins the calculations we perform, ensuring that the heat lost by the zinc matches the heat gained by the water, as shown in this exercise.

Heat Transfer Calculations

Heat transfer calculations are used to quantify how much thermal energy is transferred between substances. These calculations involve using the formula \( q = m \cdot c \cdot \Delta T \), where:

• \( q \) is the heat transferred,
• \( m \) is the mass,
• \( c \) is the specific heat capacity,
• \( \Delta T \) is the change in temperature.

In the given problem, this formula helps us find the specific heat capacity of zinc after determining the heat energy exchanged.

First, we calculated the heat gained by the water using its known specific heat capacity. Then, using the principle of conservation of energy, we equated this to the heat lost by the zinc.

By rearranging our equation, the specific heat capacity of zinc can be isolated and calculated. This method effectively applies both core thermodynamic concepts and straightforward arithmetic to solve real-world problems.