Problem 23 - College Physics

Question

\(\cdot\) You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 28.4 \(\mathrm{N}\) . You carefully add \(1.25 \times 10^{4} \mathrm{J}\) of heat energy to the sample and find that its temperature rises 18.0 \(\mathrm{C}^{\circ} .\) What is the sample's specific heat?

Step-by-Step Solution

Answer

The specific heat of the sample is approximately 239.5 J/(kg°C).

Step 1: Understand the Variables

Start by identifying the given variables in the problem: - Heat energy added, \( Q = 1.25 \times 10^{4} \text{ J} \).- Temperature rise, \( \Delta T = 18.0 \text{ °C} \).- Weight of the sample, which is 28.4 N. Remember that weight is the gravitational force, so we need to convert it to mass to use in our formula.

Step 2: Convert Weight to Mass

Convert the weight to mass using the relationship between weight and mass: Weight (N) = Mass (kg) \( \times \) gravitational acceleration (\(g = 9.8 \text{ m/s}^2\)).\[ m = \frac{\text{Weight}}{g} = \frac{28.4}{9.8} \approx 2.90 \text{ kg} \].

Step 3: Use the Specific Heat Formula

The formula to find specific heat \( c \) is:\[ Q = mc\Delta T \]Rearrange to solve for \( c \):\[ c = \frac{Q}{m\Delta T} \].Substitute the known values into the formula:\[ c = \frac{1.25 \times 10^{4} \text{ J}}{2.90 \text{ kg} \times 18.0 \text{ °C}} \].

Step 4: Calculate the Specific Heat

Now calculate the specific heat using the values:\[ c = \frac{1.25 \times 10^{4}}{2.90 \times 18.0} \approx \frac{1.25 \times 10^{4}}{52.2} \approx 239.5 \text{ J/(kg°C)} \].

Key Concepts

Heat Energy

Heat energy is a form of energy that is transferred between objects due to a temperature difference. It flows from a hotter object to a cooler one. When you add heat energy to a material, you increase its thermal energy, resulting in a change in temperature.
In the problem given, the heat energy added to the metal sample is specified as 12,500 Joules (J). This energy is absorbed by the sample, leading to an increase in its temperature. Understanding how heat energy works is crucial for determining the specific properties of materials, such as specific heat.
Additionally, it is essential to know that heat energy is measured in Joules, which is the standard unit for energy in the International System of Units (SI). Using consistent units is vital when performing calculations involving heat and temperature.

Temperature Rise

Temperature rise refers to the increase in temperature that a substance experiences when it absorbs heat energy. It is crucial to note that different materials respond differently to being heated. This response is described by the material's specific heat capacity.
In the exercise, the metal sample exhibits a temperature rise of 18.0°C when heat energy is added. This temperature change is denoted as \( \Delta T \) in formulas. This change is an important factor in calculating the specific heat of the material.
When heating any substance, the rate at which its temperature rises depends not only on the energy input but also on its mass and specific heat capacity. Hence, careful measurement of temperature changes is key in such calculations.

Mass Conversion

Mass conversion is an essential step when dealing with weight measurements. This is because weight (Newtons) must be converted to mass (kilograms) for use in most scientific equations, such as when calculating specific heat.
The problem gives the weight of the sample as 28.4 Newtons (N). To convert this to mass, we use the formula: \( m = \frac{\text{Weight}}{g} \), where \( g \) is the acceleration due to gravity (9.8 m/s²).
By inserting the provided values, we find the mass \( m \) to be approximately 2.90 kg. This mass is then a crucial parameter in calculating the specific heat of the sample.

Specific Heat Formula

The specific heat capacity of a substance is a property that describes how much heat energy is required to raise the temperature of a unit mass of the substance by one degree Celsius. This is represented by the formula:
\[ Q = mc\Delta T \]
Where:

• \( Q \) is the heat energy in Joules.
• \( m \) is the mass in kilograms.
• \( \Delta T \) is the temperature change in Celsius.
• \( c \) is the specific heat capacity in J/(kg°C).

Rearranging the formula to solve for specific heat, we have:
\[ c = \frac{Q}{m\Delta T} \]
For the given exercise values, substituting into the formula provides:
\[ c = \frac{1.25 \times 10^{4} \text{ J}}{2.90 \text{ kg} \times 18.0 \text{ °C}} \approx 239.5 \text{ J/(kg°C)} \]
Understanding this formula allows you to determine how different materials will react to heat energy and is fundamental in scientific and engineering calculations involving thermal systems.