Problem 9 - Chemistry and Chemical Reactivity

Question

The initial temperature of a 344 -g sample of iron is \(18.2^{\circ} \mathrm{C} .\) If the sample absorbs \(2.25 \mathrm{kJ}\) of energy as heat, what is its final temperature?

Step-by-Step Solution

Answer

The final temperature is approximately 32.8°C.

Step 1: Understand the given values

We have a 344 g sample of iron with an initial temperature of \( 18.2^{\circ} \mathrm{C} \). It absorbs \( 2.25 \mathrm{kJ} \) of energy as heat. We need to determine the final temperature.

Step 2: Convert energy from kJ to J

Energy is given in kilojoules (kJ), but the specific heat capacity is in Joules (J). Convert the heat energy absorbed from kJ to J: \[ 2.25 \text{ kJ} = 2.25 \times 1000 = 2250 \text{ J} \]

Step 3: Use the specific heat formula

The formula for heat transfer is: \[ q = mc\Delta T \]where \( q \) is the heat energy (2250 J), \( m \) is the mass of the substance (344 g), \( c \) is the specific heat capacity of iron (\( 0.449 \text{ J/g°C} \)), and \( \Delta T \) is the change in temperature.

Step 4: Rearrange the formula to solve for ΔT

Rearrange the formula to solve for \( \Delta T \):\[ \Delta T = \frac{q}{mc} \]Substituting the known values:\[ \Delta T = \frac{2250 \text{ J}}{344 \text{ g} \times 0.449 \text{ J/g°C}} \]

Step 5: Calculate ΔT

Calculate the change in temperature:\[ \Delta T = \frac{2250}{154.256} \approx 14.58^{\circ} \text{C} \]

Step 6: Determine the final temperature

To find the final temperature \( T_f \), add the change in temperature \( \Delta T \) to the initial temperature:\[ T_f = 18.2^{\circ} \text{C} + 14.58^{\circ} \text{C} = 32.78^{\circ} \text{C} \]

Key Concepts

Heat Transfer

Heat transfer is all about the movement of thermal energy from one object to another. In this context, we're looking at how heat energy is absorbed by an iron sample, causing its temperature to rise. This transfer of energy is why a cold pot on a stove eventually gets hot.
Understanding heat transfer is vital for solving problems where temperature changes occur due to energy absorption or release. There are three main modes of heat transfer:

• Conduction: Transfer of heat through a solid material, like a metal rod heating up when one end is placed in a fire.
• Convection: Movement of heat through fluids, such as boiling water where warm water rises and cool water sinks.
• Radiation: Energy transferred through electromagnetic waves, like heat from the sun warming your face.

In this exercise, we focus mainly on conduction, as the solid iron absorbs heat energy causing an increase in temperature.

Temperature Change

When an object absorbs heat, its temperature usually increases, depending on the material's properties. These properties are governed by factors like mass, specific heat capacity, and the energy absorbed.
To calculate the change in temperature (\( \Delta T \)), we use the equation:
\[ q = mc\Delta T \]Here, \( q \) is the amount of heat absorbed, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) represents the change in temperature. Rearranging this equation helps us find \( \Delta T \):
\[ \Delta T = \frac{q}{mc} \]By applying this formula, we can determine how much the temperature of the iron sample increases after absorbing 2.25 kJ of energy.
Knowing that the specific heat capacity of iron is relatively low means it doesn't take a lot of energy for iron to increase in temperature compared to water, for example.

Energy Conversion

Energy conversion is a fundamental concept, explaining how energy changes from one form to another. In this exercise, we're specifically converting heat energy, often measured in Joules, to a tangible change in temperature.
To work with the problem's given values efficiently, we converted 2.25 kJ of energy into Joules (J). Most scientific calculations require energy to be in Joules to align with sensible units for specific heat capacity.

• This conversion is crucial because it ensures consistency across calculations.
• 1 kJ is equivalent to 1000 J, so we multiply by 1000 to convert the units.

By understanding these conversions, we can solve various physics problems, particularly those involving heat and temperature. This reinforces that energy can be transformed across different forms and quantities, highlighting its versatility in thermodynamics and everyday scenarios.