Question

In a constant-pressure calorimetry experiment, a reaction gives off \(21.8 \mathrm{~kJ}\) of heat. The calorimeter contains \(150 \mathrm{~g}\) of water, initially at \(23.4^{\circ} \mathrm{C}\). What is the final temperature of the water? The heat capacity of the calorimeter is negligibly small.

Short answer

The final temperature of the water is \(58.23^{\circ}\text{C}\).

Step-by-step solution

Understanding the Problem

We are given that a reaction releases heat, specifically \(21.8 \text{kJ}\), and this energy is absorbed by water. We are tasked with finding the final temperature of the water.

Heat Transfer Formula

The formula for heat transfer is \( q = mc\Delta T \), where \( q \) is the heat absorbed or released, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature (\( \Delta T = T_{final} - T_{initial} \)).

Convert Units

Convert the given heat from kilojoules to joules since the specific heat capacity of water is usually given in \( \text{J}/\text{g}^\circ\text{C} \). Thus, \( q = 21.8 \text{kJ} = 21800 \text{J} \).

Rearrange Heat Transfer Formula

Rearrange the formula to solve for \( \Delta T \). We have \( \Delta T = \frac{q}{mc} \). Plug in the known values: \( q = 21800 \text{J} \), \( m = 150 \text{g} \), and \( c = 4.18 \text{J/g}^\circ\text{C} \).

Calculate Change in Temperature

Calculate \( \Delta T \) using the formula: \[ \Delta T = \frac{21800}{150 \times 4.18} \approx 34.83 \degree\text{C}. \]

Determine Final Temperature

Using \( \Delta T = T_{final} - T_{initial} \), rewrite it to solve for \( T_{final} \). Since \( T_{initial} = 23.4\degree\text{C} \), \( T_{final} = 23.4 + 34.83 = 58.23 \degree\text{C}. \)

Key concepts

heat transfer formula

Calorimetry relies on the heat transfer formula to study how heat moves between a system and its surroundings. This formula is given as \( q = mc \Delta T \). Here, \( q \) represents the heat gained or lost, \( m \) is the mass, \( c \) stands for specific heat capacity, and \( \Delta T \) indicates the temperature change from initial to final state.

Understanding this formula allows us to calculate how the temperature of an object changes when it absorbs or releases heat. It's important to note that the direction of heat flow—either into or out of the system—can affect the sign of \( q \). In this particular exercise, the negative sign indicates that the reaction gave off heat, meaning the system is losing energy rather than absorbing it.

With these components in mind, solving calorimetry problems is essentially about identifying what is known, calculating what is unknown, and applying the formula correctly. Each variable interacts with another to tell a complete story about the thermal interaction at play.

specific heat capacity

Specific heat capacity (\( c \)) is a crucial concept in calorimetry, acting as a measure of how much heat energy a substance requires to raise its temperature by one degree Celsius per gram. For water, this value is consistently \( 4.18 \, \text{J/g}^\circ \text{C} \).

This high value for water means it needs a considerable amount of energy to change its temperature, which is why water is often used as a coolant. In our calorimetry problem, water's specific heat capacity enabled us to calculate how much the water's temperature increased when it absorbed the heat from the reaction.

The larger the specific heat capacity, the more energy it takes for the substance to experience a temperature change. This property ensures that water can efficiently absorb a large amount of heat, making it perfect for experiments in constant-pressure calorimetry.

final temperature calculation

The final temperature is the culmination of our calorimetry calculations, found after determining the change in temperature, \( \Delta T \), caused by the heat transfer. We rearrange the heat transfer formula to find \( \Delta T = \frac{q}{mc} \). By substituting in \( q = 21800 \, \text{J} \), \( m = 150 \, \text{g} \), and \( c = 4.18 \, \text{J/g}^\circ \text{C} \), we get \( \Delta T \approx 34.83^\circ \text{C} \).

The initial temperature of the water was given as \( 23.4^\circ \text{C} \). So, the final temperature, \( T_{final} \), is calculated by adding the \( \Delta T \) to the initial temperature:

• Initial Temperature, \( T_{initial} = 23.4^\circ \text{C} \)
• Change in Temperature, \( \Delta T = 34.83^\circ \text{C} \)
• Final Temperature, \( T_{final} = 23.4^\circ \text{C} + 34.83^\circ \text{C} = 58.23^\circ \text{C} \)

Understanding the final temperature helps us quantify the effects of heat transfer in the reaction, providing a complete picture of the thermal transformation occurring in the system.