Problem 14 - Chemistry and Chemical Reactivity

Question

A 182 -g sample of gold at some temperature is added to 22.1 g of water. The initial water temperature is \(25.0^{\circ} \mathrm{C}\) and the final temperature is \(27.5^{\circ} \mathrm{C}\). If the specific heat capacity of gold is \(0.128 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) what was the initial temperature of the gold?

Step-by-Step Solution

Answer

The initial temperature of the gold was approximately \(37.43^{\circ}C\).

Step 1: Understand the Concept of Heat Transfer

The heat lost or gained by a substance is given by the formula \(q = m \cdot c \cdot \Delta T\), where \(q\) is the heat, \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. In this problem, the heat gained by the water will be equal to the heat lost by the gold.

Step 2: Calculate Heat Gained by Water

First, calculate the heat gained by the 22.1 g of water. Use the formula \(q = m \cdot c \cdot \Delta T\). The specific heat capacity of water is approximately \(4.18 \mathrm{J/g \cdot K}\). The temperature change \(\Delta T\) for water is \((27.5 - 25.0)^{\circ}C = 2.5^{\circ}C\). Thus, \[ q_{\text{water}} = 22.1 \cdot 4.18 \cdot 2.5 = 231.05 \text{ J} \]

Step 3: Set Heat Gained by Water Equal to Heat Lost by Gold

The heat gained by the water is equal and opposite to the heat lost by the gold. Thus, \[ q_{\text{gold}} = -q_{\text{water}} = -231.05 \text{ J} \] We are solving for the initial temperature of the gold.

Step 4: Calculate Initial Temperature of Gold

Now set up the equation for the heat lost by gold: \[ q_{\text{gold}} = 182 \cdot 0.128 \cdot (T_{\text{final}} - T_{\text{initial}}) \] We know that \(q_{\text{gold}} = -231.05 \text{ J}\) and \(T_{\text{final}} = 27.5^{\circ}C\). Thus: \[ -231.05 = 182 \times 0.128 \times (27.5 - T_{\text{initial}}) \]\[ -231.05 = 23.296 \times (27.5 - T_{\text{initial}}) \]\[ T_{\text{initial}} = 27.5 - \frac{-231.05}{23.296} \approx 37.43^{\circ}C\]

Step 5: Solve for Initial Temperature Numerically

Rearrange the final equation to find \(T_{\text{initial}}\): \[ 37.43^{\circ}C = 27.5^{\circ}C + \left(\frac{231.05}{23.296}\right) \]Therefore, the initial temperature of the gold is approximately \(37.43^{\circ}C\).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in understanding how different substances absorb or release heat energy. It is the amount of heat required to change the temperature of one gram of a substance by one degree Celsius (or Kelvin).
For different materials, the specific heat capacity can vary significantly. For instance, in this exercise, gold has a specific heat capacity of 0.128 J/g•K, indicating that it heats up or cools down with less energy compared to water.
Water, often used as a reference, has a specific heat of approximately 4.18 J/g•K. This means that water is much harder to heat or cool compared to gold.

• The lower the specific heat, the less energy is required to change the temperature.
• High specific heat substances, like water, are great at storing energy.
• Low specific heat materials, like metals, heat up and cool down quickly.

Understanding specific heat capacity helps us determine how different materials will react to temperature changes in various environmental conditions.

Temperature Change

The change in temperature, represented by \(\Delta T\), is a simple yet important component when learning about heat transfer. It indicates the difference between the final and initial temperatures of a material.
In our exercise, we see how water's temperature increases from 25.0°C to 27.5°C. This increase signifies a \(\Delta T\) of 2.5°C.
Calculating \(\Delta T\) is essential because it directly influences the amount of heat energy transferred.

• Changes in temperature are calculated by subtracting the initial temperature from the final temperature: \(\Delta T = T_{\text{final}} - T_{\text{initial}}\).
• Positive \(\Delta T\) means the substance has absorbed heat and increased in temperature.
• Negative \(\Delta T\) indicates that the substance has released heat and cooled down.

By understanding how temperature changes in a system, we can predict energy exchange events and calculate the amount of heat transferred.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It allows us to understand heat transfer in a precise way by isolating a system from its environment and measuring heat flow.
In the current exercise, we used principles of calorimetry to determine how heat transferred between water and gold. We assumed the heat lost by gold was equal to the heat gained by water, allowing us to solve for unknowns like the initial temperature of the gold.

• Calorimetry involves setting up a system where heat transfer can be measured, often using calorimeters.
• The basic formula used in calorimetry is \(q = m \cdot c \cdot \Delta T\).
• We can determine specific heat capacities or unknown temperatures in a system by using calorimetry.

By studying calorimetry, students can gain insights into how thermal energy is conserved or transferred, becoming equipped to solve complex heat-related problems.