Problem 74 - Chemistry Matter and Change

Question

How many joules of heat are lost by 3580 \(\mathrm{kg}\) of granite as it cools from \(41.2^{\circ} \mathrm{C}\) to \(-12.9^{\circ} \mathrm{C}\) ? The specific heat of granite is 0.803 \(\mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)\)

Step-by-Step Solution

Answer

The granite loses 155,668,694 joules of heat.

Step 1: Convert Mass to Grams

Since the specific heat capacity is given in joules per gram degree Celsius, we first convert the mass of granite from kilograms to grams. We know that 1 kg equals 1000 grams, so:\[ 3580 \text{ kg} = 3580 \times 1000 = 3,580,000 \text{ grams} \]

Step 2: Identify Temperature Change

Calculate the temperature change (\( \Delta T \)) by subtracting the final temperature from the initial temperature:\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = -12.9^{\circ} \text{C} - 41.2^{\circ} \text{C} \]\[ \Delta T = -54.1^{\circ} \text{C} \]

Step 3: Use the Heat Loss Formula

Utilize the formula for heat transfer, which is:\[ Q = mc\Delta T \]where \( Q \) is the heat lost, \( m \) is the mass in grams, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. Substituting the known values:\[ m = 3,580,000 \text{ grams}, \quad c = 0.803 \text{ J/g}^{\circ}\text{C}, \quad \Delta T = -54.1^{\circ}\text{C} \]\[ Q = 3,580,000 \times 0.803 \times (-54.1) \]

Step 4: Calculate the Heat Lost

Compute the value for \( Q \):\[ Q = 3,580,000 \times 0.803 \times (-54.1) \]\[ Q = -155,668,694 \text{ joules} \]The negative sign indicates that heat is lost.

Key Concepts

Understanding Specific Heat Capacity

Specific heat capacity is a fundamental concept in thermochemistry. It is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. Often represented by the symbol \( c \), it indicates how a material responds to the addition or removal of heat.

Different materials will have different specific heat capacities. For instance, water has a high specific heat capacity, which means it can store a lot of energy without a large change in temperature. In contrast, materials like metals may have lower specific heat capacities. In our example, the specific heat capacity of granite is given as 0.803 \( ext{J/g}^ ext{°C} \). This means that 0.803 joules of energy are needed to raise the temperature of 1 gram of granite by 1°C.

When determining how much heat is lost or gained during a temperature change, knowing the specific heat capacity is crucial.

Analyzing Temperature Change

Temperature change is a key part of calculating heat transfer. Often symbolized by \( \Delta T \), it is the difference between the final and initial temperatures of a material. In thermochemistry, it is crucial to determine \( \Delta T \) to find out how the state of a substance has been altered.

In our case with granite, we calculate \( \Delta T \) by subtracting the final temperature \( -12.9^{\circ} \text{C} \) from the initial temperature \( 41.2^{\circ} \text{C} \). This results in a temperature change of \(-54.1^{\circ} \text{C} \). The calculation shows that granite is cooling down.

Remember that when \( \Delta T \) is negative, like here, it denotes a decrease in temperature, leading to the possibility of heat loss.

Exploring Heat Transfer

Heat transfer describes how thermal energy moves from one place to another. In thermochemistry, it is vital to comprehend how energy is transferred, especially between different materials. The movement of heat is typically calculated using the formula \( Q = mc\Delta T \).

In this formula:

• \( Q \) represents the heat energy gained or lost (in joules).
• \( m \) is the mass of the substance (in grams).
• \( c \) is the specific heat capacity.
• \( \Delta T \) is the temperature change.

Using these variables, you can determine how much energy is transferred as a substance changes temperature. In our granite example, this formula helps calculate the joules of heat transferred as the material cools.

The Calculation of Heat Loss

Calculation of heat loss involves utilizing the known values in the heat transfer formula. It allows us to determine how much energy a substance has released as it cools.

For the granite in our example, we substitute its specific heat capacity, mass in grams, and temperature change into the equation \( Q = mc\Delta T \).

Here's how you compute it:

• Mass \( m = 3,580,000 \text{ grams} \).
• Specific heat \( c = 0.803 \text{ J/g}^\circ\text{C} \).
• Temperature change \( \Delta T = -54.1^\circ \text{C} \).

Plug these values in: \[ Q = 3,580,000 \times 0.803 \times (-54.1) \]

The calculation results in a heat loss of \(-155,668,694\) joules. The negative sign signifies that the granite released energy as it cooled.