Problem 42 - Chemistry and Chemical Reactivity

Question

A \(9.36-\) g piece of platinum is heated to \(98.6^{\circ} \mathrm{C}\) in a boiling water bath and then dropped onto ice. (See Study Question 41.) When the metal's temperature has dropped to \(0.0^{\circ} \mathrm{C},\) it is found that \(0.37 \mathrm{g}\) of ice has melted. What is the specific heat capacity of platinum?

Step-by-Step Solution

Answer

The specific heat capacity of platinum is approximately 0.134 J/g°C.

Step 1: Understanding the Problem

We need to find the specific heat capacity of platinum. A piece of platinum is heated to a specific temperature, dropped onto ice, and melts a certain amount of ice. We can use the concept of heat transfer to solve this problem.

Step 2: Conceptualizing Heat Transfer

The heat lost by the platinum as it cools is equal to the heat gained by the ice as it melts. The equation for heat transfer due to temperature change is given by \( q = mc\Delta T \). The heat absorbed by ice to melt is given by \( q = mL_f \), where \( L_f \) is the latent heat of fusion.

Step 3: Setting Up the Equations

The energy lost by platinum \( q_p \) is \( mc\Delta T = 9.36 \text{ g} \times c \times (98.6 - 0.0) \text{ °C} \). The energy gained by ice is \( q_i = 0.37 \text{ g} \times 334 \text{ J/g} \) (latent heat of fusion of ice is 334 J/g). Since \( q_p = q_i \), set the equations equal: \( 9.36c \times 98.6 = 0.37 \times 334 \).

Step 4: Solving for Specific Heat Capacity

Solve the equation for \( c \), the specific heat capacity of platinum. First calculate the heat absorbed by ice: \( 0.37 \times 334 = 123.58 \text{ J} \). Then, rearrange \( 9.36c \times 98.6 = 123.58 \) to find \( c \): \( c = \frac{123.58}{9.36 \times 98.6} \).

Step 5: Calculate the Solution

Compute the value: \( c = \frac{123.58}{922.896} \approx 0.134 \text{ J/g°C} \). This is the specific heat capacity of platinum.

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from one system to another due to a temperature difference. In this exercise, the hot platinum is placed on ice. Here, heat flows from the warmer platinum to the colder ice. As a result, the platinum cools down and the ice melts.

We can understand this process using the concept of heat conservation. The heat lost by the platinum equals the heat gained by the ice. This means all energy leaving the platinum is the same as the energy turning the ice into water. This is crucial in calculating specific heat capacity because it shows the energy balance between both objects.

• Platinum loses energy (cools down)
• Ice gains energy (melts into water)

The mathematical expression for heat transfer when a substance changes temperature is: \( q = mc\Delta T \), where

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

This formula helps us see how much thermal energy is transferred from one body to another.

Latent Heat of Fusion

Latent heat of fusion is the energy needed to change a substance from solid to liquid without a temperature change. When ice melts, it requires a specific amount of energy, even though its temperature does not increase. This energy is called the latent heat of fusion.

In our problem, when the platinum is put on the ice, some of the ice melts. This indicates that the energy transfer from the platinum is used to overcome the solid structure of ice, turning it into water. We can find the energy required for this phase change using the formula \( q = mL_f \), where:

• \( q \) is the heat energy required
• \( m \) is the mass of the ice melted
• \( L_f \) is the latent heat of fusion (for ice, it is \( 334 \text{ J/g} \))

This principle explains why, despite the temperature being the same at \( 0^{\circ} \text{C} \), ice requires energy to melt, which is derived from the cooling platinum.

Heat Calculation

To solve problems involving heat transfer, we often use heat calculations. These involve determining the energy exchanged between substances. In this exercise, the final goal is to find the specific heat capacity of platinum through a series of calculations.

Once we know the energy used to melt the ice (using the latent heat of fusion), we can equate it to the heat lost by platinum. The equation set up from energy balance is critical:

• Heat lost by platinum: \( q_p = mc\Delta T = 9.36 \times c \times (98.6 - 0.0) \)
• Heat gained by ice: \( q_i = 0.37 \times 334 \)

We equate these values because energy must be conserved, \( q_p = q_i \). This allows us to solve for the specific heat capacity, \( c \), with the formula:

• \( c = \frac{q_i}{m\Delta T} \)

Plugging the numbers gives us the specific heat capacity of platinum. This calculation shows how heat transfer analysis can help in finding properties of materials.