Problem 64 - General Chemistry

Question

An iron skillet weighing \(1.28 \mathrm{~kg}\) is heated on a stove to \(178^{\circ} \mathrm{C}\). Suppose the skillet is cooled to room temperature, \(21^{\circ} \mathrm{C}\). How much heat energy (in joules) must be removed to effect this cooling? The specific heat of iron is \(0.449 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).

Step-by-Step Solution

Answer

90,270.56 joules of heat energy must be removed.

Step 1: Convert Units

First, convert the weight of the skillet from kilograms to grams because the specific heat capacity is given in terms of grams. Since there are 1000 grams in a kilogram, we have: Weight of skillet = \(1.28 \text{ kg} \times 1000 \frac{\text{grams}}{\text{kg}} = 1280 \text{ grams}\).

Step 2: Identify Initial and Final Temperatures

Next, identify the initial and final temperatures. The initial temperature \(T_i\) is \(178^{\circ} \text{C}\), and the final temperature \(T_f\) is \(21^{\circ} \text{C}\).

Step 3: Calculate Temperature Change

Calculate the change in temperature \(\Delta T\) by subtracting the final temperature from the initial temperature:\[\Delta T = T_i - T_f = 178^{\circ} \text{C} - 21^{\circ} \text{C} = 157^{\circ} \text{C}\]

Step 4: Use the Heat Energy Formula

The amount of heat energy \(Q\) required to change the temperature of a substance can be calculated using the formula: \[Q = m \cdot c \cdot \Delta T\]where:- \(m\) is the mass of the skillet (1280 grams),- \(c\) is the specific heat capacity of iron (0.449 \text{ J/g°C}), and- \(\Delta T\) is the temperature change (157°C).

Step 5: Compute the Heat Energy

Substitute the values into the formula:\[Q = 1280 \text{ grams} \cdot 0.449 \text{ J/g°C} \cdot 157^{\circ} \text{C} = 90270.56 \text{ J}\]Thus, 90270.56 joules of heat energy must be removed to cool the skillet from \(178^{\circ} \text{C}\) to \(21^{\circ} \text{C}\).

Key Concepts

Specific Heat Capacity

The specific heat capacity is a property of matter that reflects how much heat energy a substance can store per unit of mass for a given temperature change. It is usually expressed in joules per gram per degree Celsius (J/g°C). For iron, the specific heat capacity is 0.449 J/g°C.

This means that for every gram of iron, 0.449 joules of energy will change its temperature by one degree Celsius. A higher specific heat capacity means a material can absorb more heat for the same temperature change, while a lower value indicates it heats up or cools down faster with less energy involved. Understanding the specific heat capacity is crucial for calculating the energy involved in heating or cooling processes in different materials.

In this case of an iron skillet, knowing the specific heat capacity allows us to determine how much heat energy must be removed to lower the temperature of the skillet.

Temperature Change

Temperature change (\(\Delta T\)) is the difference between the initial temperature (\(T_i\)) and the final temperature (\(T_f\)) of an object. It is important in calculating heat energy because it shows how much energy is needed to change the object's temperature.

To find \(\Delta T\), simply subtract the final temperature from the initial temperature: \[\Delta T = T_i - T_f\]
In the exercise, the skillet initially at 178°C cools down to 21°C, resulting in a temperature change of 157°C. This information, combined with the specific heat capacity, helps determine the total heat energy involved in the cooling process.

Grasping the concept of temperature change is essential since it directly impacts the heat energy calculations in processes like heating, cooling, or even mixing materials.

Iron

Iron is a common metal with some unique properties that make it widely used in cooking utensils like skillets. It has a specific heat capacity of 0.449 J/g°C, which indicates how it behaves in heating and cooling processes.

Iron can store a moderate amount of heat, which means it will heat up relatively quickly and evenly, making it ideal for cooking. It is not as quick to change temperature as some other metals, like aluminum, due to its specific heat capacity. However, once iron heats up, it maintains its temperature well, which is useful for cooking processes that benefit from consistent heat.

Understanding the characteristics of iron helps in practical applications like determining the energy requirements for heating and cooling, as well as forming expectations about how quickly and effectively a skillet will respond to temperature changes.

Cooling Process

The cooling process is the method of extracting heat from an object to lower its temperature. For an object like an iron skillet, the process involves removing the heat it absorbed during heating.

Using the formula: \[Q = m \cdot c \cdot \Delta T\]
we can calculate the amount of heat energy (\(Q\)) that needs to be removed to achieve the desired temperature decrease. Here:

• \(m\) is the mass of the object (1280 grams for our skillet)
• \(c\) is the specific heat capacity (0.449 J/g°C for iron)
• \(\Delta T\) is the temperature change (157°C in this example)

Combining these values gives us the total energy required for cooling the skillet to room temperature. This calculation is crucial in understanding and managing energy use in various practical and industrial applications involving temperature control.