Problem 5 - General Chemistry: Principles and Modern Applications

Question

A 465 g chunk of iron is removed from an oven and plunged into \(375 \mathrm{g}\) water in an insulated container. The temperature of the water increases from 26 to \(87^{\circ} \mathrm{C}\). If the specific heat of iron is \(0.449 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1},\) what must have been the original temperature of the iron?

Step-by-Step Solution

Answer

The original temperature of the iron was approximately 544.7 degrees Celsius.

Step 1: Compute the Heat Gained by the Water

Use the formula \( Q = m \cdot C \cdot \Delta T \) to calculate the heat gained by the water, where \( Q \) is the heat, \( m \) is the mass, \( C \) is the specific heat, and \( \Delta T \) is the temperature change. For water, \( m = 375g \), \( C = 4.18 Jg^{-1} ^{\circ} C^{-1} \) (the specific heat of water), and \( \Delta T = 87^{\circ}C - 26^{\circ}C = 61^{\circ}C \). Substituting these values we get \( Q = 375g \cdot 4.18 Jg^{-1} ^{\circ} C^{-1} \cdot 61^{\circ}C = 95722.5 J \).

Step 2: Compute the Temperature Change of the Iron

The heat lost by the iron is equal to the heat gained by the water, so \( Q_{iron} = Q_{water} = 95722.5 J \). We can rearrange the formula from step 1 to find the temperature change of the iron as \( \Delta T = Q / (m \cdot C) \). The mass of the iron is given as 465g and the specific heat is 0.449 Jg^{-1} ^{\circ} C^{-1}, so the temperature change is \( \Delta T = 95722.5 J / (465g \cdot 0.449 Jg^{-1} ^{\circ} C^{-1}) = -457.7^{\circ}C \). The negative sign indicates that the temperature of the iron decreased.

Step 3: Compute the Original Temperature of the Iron

To find the original temperature of the iron, add the magnitude of the temperature change to the final temperature. The final temperature of the iron is the same as the final temperature of the water, which is 87^{\circ}C. So the original temperature of the iron is \( 87^{\circ}C + 457.7^{\circ}C = 544.7^{\circ}C \).

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from one object to another due to a difference in temperature. In our scenario, heat transfer occurs when the hot iron is submerged into water, causing the heat from the iron to transfer to the water. This happens because the iron and water reach thermal equilibrium, meaning they share the same energy temperature.

Here, the heat energy moves away from the iron to the water until both have a uniform temperature. The principle is that energy cannot be created or destroyed—only transformed. This means during heat transfer, the amount of heat lost by the iron is equal to the amount of heat gained by the water. The calculation of this heat exchange involves the formula:

• \( Q = m \cdot C \cdot \Delta T \)

where \( Q \) represents the heat transfer, \( m \) is the mass, \( C \) is the specific heat, and \( \Delta T \) is the change in temperature. Understanding this equation allows us to solve many problems involving thermal energy and temperature changes.

Specific Heat

Specific heat is a measure of how much energy is needed to raise the temperature of a particular substance. Every material has a specific heat value, which indicates how well it can store heat energy compared to other materials. In this exercise, specific heat helps us understand how much energy is required for the iron and water to change temperature.

Iron has a specific heat of \( 0.449 \, \mathrm{Jg}^{-1} \, ^{\circ} \mathrm{C}^{-1} \), while water has a specific heat of \( 4.18 \, \mathrm{Jg}^{-1} \, ^{\circ} \mathrm{C}^{-1} \). The higher specific heat of water means it needs more energy to increase its temperature compared to iron. In the exercise, the specific heat of these substances helps determine the amount of heat needed to change their temperatures. This is why the vast amount of energy transferred from iron is reflected in the significant temperature increase of the water compared to its mass.

Temperature Change

Temperature change is an essential part of calorimetry, which focuses on measuring how substances exchange heat. In the context of calorimetry, temperature change tells us how much a substance's temperature varies due to heat transfer.

For water, its temperature increased by \( 61^{\circ} \mathrm{C} \) (from \( 26^{\circ} \mathrm{C} \) to \( 87^{\circ} \mathrm{C} \)), which allows us to calculate the energy gained from the iron. For iron, the heat loss results in a temperature drop calculated using its heat loss and specific heat. The iron's temperature change is significant, \( -457.7^{\circ} \mathrm{C} \), indicating a large temperature difference.

Knowing the temperature change of each material is crucial in determining the initial temperatures before heat exchange and assessing how rapidly and significantly one material loses or gains heat. Using the equation \( \Delta T = Q / (m \cdot C) \), we can calculate this change accurately by working backwards from the known heat energy and specific data on the materials involved.