Problem 23 - College Physics

Question

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 \( 239.6 \mathrm{~J/(kg \cdot C^{\circ})} \).

Step 1: Understanding the Given Values

We are given the following: the weight of the sample, \( W = 28.4 \mathrm{~N} \), the heat added, \( Q = 1.25 \times 10^4 \mathrm{~J} \), and the change in temperature, \( \Delta T = 18.0 \mathrm{~C}^{\circ} \). We need to find the specific heat \( c \) of the sample.

Step 2: Calculating the Mass

First, convert weight (force) to mass using the formula \( W = mg \) where \( g = 9.8 \mathrm{~m/s^2} \). Rearrange to find \( m = \frac{W}{g} = \frac{28.4 \mathrm{~N}}{9.8 \mathrm{~m/s^2}} \approx 2.9 \mathrm{~kg} \).

Step 3: Using the Specific Heat Formula

The formula for specific heat is \( c = \frac{Q}{m \Delta T} \). Substitute the respective values: \( Q = 1.25 \times 10^4 \mathrm{~J} \), \( m = 2.9 \mathrm{~kg} \), and \( \Delta T = 18.0 \mathrm{~C}^{\circ} \) into the equation.

Step 4: Calculating Specific Heat

Put the values into the formula: \[ c = \frac{1.25 \times 10^4 \mathrm{~J}}{2.9 \mathrm{~kg} \times 18.0 \mathrm{~C}^{\circ}} \approx 239.6 \mathrm{~J/(kg \cdot C^{\circ})} \].

Step 5: Verifying Units

Check that the final units of specific heat are \( \mathrm{J/(kg \cdot C^{\circ})} \), which are the correct and standard units for specific heat capacity in the International System of Units (SI).

Key Concepts

Specific Heat

Specific heat is a fundamental concept in physics that describes how much energy is needed to change the temperature of a unit mass of a substance by one degree. It's symbolized as \(c\) and expressed in the units \(\mathrm{J/(kg \cdot C^{\circ})}\).
Specific heat varies depending on the material. This means different materials require different amounts of energy to change their temperatures.

• Higher specific heat means more energy is needed for temperature change.
• Lower specific heat indicates less energy needed.

This property is crucial in understanding how substances heat up and cool down in our everyday environment.

Heat Energy

Heat energy, often represented as \(Q\), is the energy transferred, typically as a result of temperature differences. In physics, it is measured in joules (\(\mathrm{J}\)).

• Heat can be transferred by conduction, convection, and radiation.
• It's a form of energy that can cause changes in the temperature of substances.

During our exercise, the metal sample absorbed \(1.25 \times 10^4\) joules of heat. Understanding how this heat energy affects the metal's temperature is key to solving the problem of specific heat.

Temperature Change

Temperature change, denoted as \(\Delta T\), is the difference in the initial and final temperatures of a substance after heat has been applied. It is measured in degrees Celsius (\(\mathrm{C^{\circ}}\)).
This concept is a direct indicator of the effect of heat energy on a substance.

• A small temperature change might suggest a high specific heat.
• A large temperature change often indicates a lower specific heat.

In our metal sample exercise, the temperature increased by \(18.0 \mathrm{C^{\circ}}\). Calculating this change helps us better estimate the specific heat when combined with other measurements.

Physics Problem Solving

Physics problem solving requires a structured approach to understand and resolve physical phenomena and their mathematical representations. In our exercise, this involves several key steps:

• Identifying given values (weight, heat energy, temperature change).
• Converting weight to mass, using the relationship between force and mass \(W = mg\).
• Applying the specific heat formula \(c = \frac{Q}{m \Delta T}\).
• Substituting known values into the formula to compute the result.

This systematic approach ensures accuracy and understanding of the physical concepts involved. By organizing and calculating each component step-by-step, like we did, you can effectively solve similar physics problems in the future.