Problem 17 - 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 come to thermal equilibrium, the temperature is \(27.1^{\circ} \mathrm{C} .\) What is the specific heat capacity of zinc?

Step-by-Step Solution

Answer

The specific heat capacity of zinc is approximately 0.39 J/g°C.

Step 1: Identify the Known Values

We are given the mass of zinc, which is 13.8 g, its initial temperature of 98.8°C, and the equilibrium temperature of 27.1°C. We are also given the mass of water, 45.0 g, its initial temperature of 25.0°C, and its specific heat capacity of 4.18 J/g°C.

Step 2: Write the Heat Transfer Equations

Use the formula for heat transfer: \[ q = m \cdot c \cdot \Delta T \]where \(q\) is the heat absorbed or released, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

Step 3: Set the Equation for Heat Loss by Zinc

The heat lost by zinc is given by:\[ q_{\text{zinc}} = m_{\text{zinc}} \cdot c_{\text{zinc}} \cdot (T_{\text{final}} - T_{\text{initial,zinc}}) \]where \(m_{\text{zinc}} = 13.8 \) g, \(T_{\text{final}} = 27.1\)°C, and \(T_{\text{initial,zinc}} = 98.8\)°C.

Step 4: Set the Equation for Heat Gain by Water

The heat gained by water is given by:\[ q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial,water}}) \]where \(m_{\text{water}} = 45.0 \) g, \(c_{\text{water}} = 4.18 \) J/g°C, and \(T_{\text{initial,water}} = 25.0\)°C.

Step 5: Set the Net Heat Exchange to Zero

Since no heat is lost to the surroundings, the heat lost by zinc equals the heat gained by water:\[ m_{\text{zinc}} \cdot c_{\text{zinc}} \cdot (27.1 - 98.8) = m_{\text{water}} \cdot c_{\text{water}} \cdot (27.1 - 25.0) \]

Step 6: Solve for Specific Heat Capacity of Zinc

Plug the known values into the equation and solve for \(c_{\text{zinc}}\):\[ 13.8 \cdot c_{\text{zinc}} \cdot (-71.7) = 45.0 \cdot 4.18 \cdot (2.1) \]Calculate:\[ 13.8 \cdot c_{\text{zinc}} \cdot (-71.7) = 396.57 \]\[ c_{\text{zinc}} = \frac{396.57}{13.8 \cdot (-71.7)} \approx 0.39 \, \text{J/g°C} \]

Step 7: Conclusion

The specific heat capacity of zinc is approximately 0.39 J/g°C.

Key Concepts

Thermal Equilibrium

When an object, like a piece of zinc, is heated and then placed in water, they eventually reach a point where they are both at the same temperature. This state is known as thermal equilibrium. At thermal equilibrium, the heat lost by the zinc equals the heat gained by the water, assuming no heat is lost to the surrounding. This is why thermal equilibrium is an essential concept in understanding heat transfer.

• Thermal equilibrium occurs when the temperatures of two substances become equal.
• Assuming no heat loss to surroundings, energy transfer between the substances balances out.

Reaching thermal equilibrium is crucial for solving problems involving specific heat. It allows us to use all given temperatures and specific heat capacities to find unknowns like the specific heat of zinc in this exercise.

Heat Transfer

Heat transfer is the movement of thermal energy from one substance to another, and it can occur in various ways. In this exercise, we are dealing with heat transfer through direct contact, where the zinc and water exchange heat until they reach thermal equilibrium.

• Heat moves from the hotter object (zinc) to the cooler one (water).
• The formula for heat transfer is: \( q = m \cdot c \cdot \Delta T \).

This formula helps calculate how much heat is exchanged:

• \( q \) is the heat transferred (in joules);
• \( m \) is the mass of the substance (zinc or water);
• \( c \) is the specific heat capacity;
• \( \Delta T \) is the temperature change.

By equating heat lost by zinc to heat gained by water, we preserve energy conservation and solve for unknown quantities.

Specific Heat of Water

The specific heat of a substance tells us how much heat is required to change its temperature. The specific heat of water is quite high compared to many other substances. It is precisely 4.18 J/g°C, indicating that water can absorb significant amounts of heat with minimal temperature change. This property makes water an excellent medium for heat exchange in experiments.

• Specific heat describes the heat needed to raise 1 gram of a substance by 1°C.
• Water's high specific heat helps regulate temperatures and support life's thermal stability.

In our exercise, this property of water is pivotal. It enables us to apply the heat transfer formula effectively and solve for zinc's specific heat capacity.