Problem 21 - Chemistry Matter and Change

Question

Calculate the specific heat in \(\mathrm{J} /\left(\mathrm{q} \cdot^{\circ} \mathrm{C}\right)\) of an unknown substance if a \(2.50-\mathrm{g}\) sample releases 12.0 cal as its temperature changes from \(25.0^{\circ} \mathrm{C}\) to \(20.0^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The specific heat of the substance is approximately 4.02 J/g°C.

Step 1: Determine the heat released in Joules

The problem provides the heat released in calories, but we need it in joules for the specific heat. Use the conversion factor, where 1 calorie (cal) equals 4.184 Joules (J). Therefore, \( 12.0 \text{ cal} = 12.0 \times 4.184 \text{ J} = 50.208 \text{ J} \).

Step 2: Calculate the Temperature Change

Determine the change in temperature (\( \Delta T \)) of the substance. The initial temperature (\( T_i \)) is \( 25.0^{\circ}\text{C} \) and the final temperature (\( T_f \)) is \( 20.0^{\circ}\text{C} \). Therefore, \( \Delta T = T_i - T_f = 25.0 - 20.0 = 5.0^{\circ}\text{C} \).

Step 3: Use the Formula for Specific Heat

The formula for specific heat capacity is \( c = \frac{q}{m \cdot \Delta T} \), where \( q \) is the heat released in joules, \( m \) is the mass in grams, and \( \Delta T \) is the temperature change. Substitute the known values: \( q = 50.208 \text{ J} \), \( m = 2.50 \text{ g} \), and \( \Delta T = 5.0^{\circ} \text{C} \).

Step 4: Solve for Specific Heat

Substitute the values into the formula: \[ c = \frac{50.208 \text{ J}}{2.50 \text{ g} \times 5.0^{\circ}\text{C}} = \frac{50.208}{12.5} \approx 4.01664 \text{ J/g}^{\circ}\text{C} \. \] Round the result as needed for significant figures.

Key Concepts

Calories to Joules Conversion

When dealing with thermodynamics, it is often necessary to convert units for accurate calculations. Calories and joules are two common units for measuring energy or heat. One calorie is defined as the amount of energy needed to raise the temperature of 1 gram of water by 1 degree Celsius. Meanwhile, joules are part of the International System of Units (SI). To convert calories to joules, we use the conversion factor: 1 calorie equals 4.184 joules.
Thus, to convert 12.0 calories to joules, multiply 12.0 by 4.184, resulting in approximately 50.208 joules. This conversion is crucial because scientific calculations often require consistency in using SI units. This ensures our calculations align with standard scientific practices.

• 1 cal = 4.184 J
• 12.0 cal = 12.0 × 4.184 J = 50.208 J

Temperature Change

Temperature change (\( \Delta T \)) is the difference between the initial and final temperatures of a substance during a cooling or heating process. Understanding this change is essential to determining how much heat has been transferred within a system.
In the given problem, the initial temperature (\( T_i \)) is 25.0°C, and the final temperature (\( T_f \)) is 20.0°C. To calculate the temperature change, subtract the final temperature from the initial temperature:

• \( \Delta T = T_i - T_f = 25.0 - 20.0 = 5.0°C \)

This calculation shows that the unknown substance's temperature decreased by 5.0°C, indicating energy release as heat.

Specific Heat Formula

The specific heat formula is a fundamental equation in thermodynamics, expressing how much energy is required to change the temperature of a given mass. The specific heat capacity (\( c \)) of a material tells us how much heat energy is needed to raise 1 gram of the substance by 1 degree Celsius.
The formula is expressed as:

• \( c = \frac{q}{m \cdot \Delta T} \)

where:

• \( q \) is the heat absorbed or released (in joules),
• \( m \) is the mass of the substance (in grams),
• \( \Delta T \) is the temperature change (in degrees Celsius).

By substituting known values into this formula, we can solve for the specific heat capacity of the substance, offering valuable insights into its thermal properties.

Heat Released

Heat released (\( q \)) refers to the amount of energy emitted as a substance cools down or a reaction occurs. This concept is intrinsic to understanding how energy transfers in physical or chemical changes.
In the exercise, the substance releases 12.0 calories of heat as it cools from 25.0°C to 20.0°C. By converting this energy to joules for specificity in calculations, we find that 50.208 joules of heat are released.
This energy release is crucial in computations and allows us to apply the specific heat formula to identify the material characteristics.

Mass in Grams

Mass is a measure of the amount of matter in a given sample, often expressed in grams for laboratory and scientific purposes. Understanding the mass of a substance is necessary for calculating specific heat since it directly influences the amount of heat absorbed or released during temperature change.
In our problem, the mass of the unknown substance is 2.50 grams. This value is vital when using the specific heat formula to determine how much energy is necessary to change the temperature of one gram of the substance by one degree Celsius.
By knowing the mass, we can accurately apply the specific heat formula and understand the material's thermal properties and how it interacts with thermal energy.