Problem 101 - Chemistry Matter and Change

Question

The temperature of 756 \(\mathrm{g}\) of water in a calorimeter increases from \(23.2^{\circ} \mathrm{C}\) to \(37.6^{\circ} \mathrm{C} .\) How much heat was given off by the reaction in the calorimeter? (Chapter 15\()\)

Step-by-Step Solution

Answer

The heat given off by the reaction in the calorimeter is 45.36 kJ, which was calculated using the formula \(q = mc\Delta T\) and the given values for mass, specific heat capacity of water, and the temperature change.

Step 1: Calculate the change in temperature

To find the change in temperature (ΔT), subtract the initial temperature (\(T_{initial}\)) from the final temperature (\(T_{final}\)): ΔT = \(T_{final}\) - \(T_{initial}\) ΔT = 37.6°C - 23.2°C ΔT = 14.4°C

Step 2: Use the heat formula to calculate the heat

Now, use the formula \(q = mc\Delta T\) to determine the heat given off by the reaction: q = (756 g) × (4.18 J/g°C) × (14.4°C)

Step 3: Calculate the heat

Multiply the values from step 2: q = 756 × 4.18 × 14.4 q = 45362.88 J

Step 4: Express the answer in kilojoules

To express the heat in kilojoules (kJ), divide the result by 1000: q = 45362.88 J ÷ 1000 q = 45.36 kJ The heat given off by the reaction in the calorimeter is 45.36 kJ.

Key Concepts

Heat Calculation

When we're tasked with figuring out how much heat is absorbed or released in a chemical reaction, we use the concept of heat calculation. In calorimetry, this is vital for determining the energy changes that occur during reactions in a calorimeter.

To calculate the heat released or absorbed, we use the formula:

• \( q = mc\Delta T \)

where:

• \( q \) is the heat energy exchanged (in Joules \(J\) or kilojoules \(kJ\))
• \( m \) is the mass of the substance (in grams \(g\))
• \( c \) is the specific heat capacity
• \( \Delta T \) is the change in temperature

To determine \( q \), we multiply the mass of the substance by its specific heat and the change in temperature. Applying this formula allows us to quantify how much heat is involved in raising or lowering the temperature of a substance, making it a fundamental step in calorimetry analysis.

Temperature Change

One of the key elements in calorimetry is understanding how temperature change affects the heat calculations. Temperature change, denoted by \( \Delta T \), is simply the difference in temperature from the start to the end of a reaction or process.

The formula for temperature change is:

• \( \Delta T = T_{final} - T_{initial} \)

Where:

• \( T_{final} \) represents the final temperature of the system
• \( T_{initial} \) is the initial temperature

In our exercise, the water in the calorimeter experienced a temperature increase from 23.2°C to 37.6°C. Calculating this:

• \( \Delta T = 37.6^{\circ} \mathrm{C} - 23.2^{\circ} \mathrm{C} = 14.4^{\circ} \mathrm{C} \)

Knowing this change is crucial because it directly influences the amount of heat calculated using the heat formula. When \( \Delta T \) is positive, it indicates a temperature rise, meaning heat is absorbed by the system.

Specific Heat Capacity

Specific heat capacity is an essential concept in calorimetry. It refers to the amount of heat needed to change the temperature of one gram of a substance by one degree Celsius.

Specific heat capacity, denoted as \( c \), plays a central role in determining how different substances respond to heat. It's a constant that varies depending on the material.

• For water, \( c = 4.18 \) J/g°C

This means to raise the temperature of one gram of water by 1°C, 4.18 Joules of energy are required. Knowing the specific heat capacity of a substance allows us to predict how much it heats up or cools down when energy is transferred.

For the calculation in our exercise, we used the specific heat capacity of water because the reaction occurs in an aqueous environment. By understanding and applying the specific heat capacity, students can accurately relate energy changes to the resulting changes in temperature during chemical reactions or physical processes.