Chapter 11: Problem 8
A \(5.00-g\) pellet of aluminum reaches a final temperature of \(63^{\circ} \mathrm{C}\) when gaining \(200 \mathrm{~J}\) of heat. What is its initial temperature?
Short Answer
Expert verified
The initial temperature of the aluminum pellet is approximately 18.56°C.
Step by step solution
01
Understand the Heat Transfer Formula
The heat transfer formula that relates the heat gained/lost by a substance is given by the equation \( q = mc\Delta T \), 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.
02
Identify Known Values
From the problem, we know: the mass \( m = 5.00 \) g, the final temperature \( T_f = 63^{\circ} \mathrm{C} \), and the heat gained \( q = 200 \mathrm{~J} \). The specific heat capacity of aluminum is approximately \( c = 0.900 \frac{\mathrm{J}}{\mathrm{g}^{\circ} \mathrm{C}} \).
03
Rearrange Equation to Solve for Initial Temperature
The change in temperature \( \Delta T \) is the difference between the final and initial temperatures \( T_f - T_i \). Rearrange the formula to solve for \( T_i \): \[ T_i = T_f - \frac{q}{mc} \].
04
Substitute Values into Equation
Substitute the known values into the rearranged formula: \[ T_i = 63^{\circ} \mathrm{C} - \frac{200 \mathrm{~J}}{5.00 \mathrm{~g} \times 0.900 \frac{\mathrm{J}}{\mathrm{g}^{\circ} \mathrm{C}}} \].
05
Calculate Initial Temperature
Perform the calculation: \[ \frac{200}{5.00 \times 0.900} = \frac{200}{4.50} \approx 44.44 \]. Thus, \( T_i = 63 - 44.44 \) which gives approximately \( 18.56^{\circ} \mathrm{C} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Specific Heat Capacity
Specific heat capacity is a crucial concept when studying heat transfer. It tells us the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius. In more straightforward terms, it measures how much heat energy a substance can hold before its temperature changes.
Here's how it works in practice:
Here's how it works in practice:
- Substances with high specific heat capacity can absorb a lot of heat without much change in temperature. Water is a classic example, which is why it's used in cooling systems.
- Conversely, materials with low specific heat capacity heat up and cool down quickly.
Exploring Temperature Change
Temperature change, denoted as \( \Delta T \), represents the difference between the final temperature and the initial temperature of a substance after absorbing or losing heat. In any scenario involving heat transfer, knowing the temperature change is key to understanding the energy exchange involved.
In the formula \( q = mc\Delta T \), \( \Delta T \) is calculated by subtracting the initial temperature \( T_i \) from the final temperature \( T_f \). That calculation helps determine how much heat energy was transferred.
In the formula \( q = mc\Delta T \), \( \Delta T \) is calculated by subtracting the initial temperature \( T_i \) from the final temperature \( T_f \). That calculation helps determine how much heat energy was transferred.
- Increases in temperature typically indicate that the substance absorbed heat (endothermic process).
- Decreases in temperature suggest the substance lost heat (exothermic process).
Investigating Aluminum Pellets
Aluminum is a lightweight metal often used in thermal experiments due to its well-defined heat-transfer properties. Its specific heat capacity is lower compared to water, which means it heats up relatively quickly.
- This makes aluminum pellets ideal for studying quick temperature variations when heat is applied or removed.
- These pellets are also commonly used in educational settings to illustrate heat transfer principles. Their size, shape, and heat capacity make them manageable and straightforward to work with in experiments.