Problem 26 - General Chemistry

Question

A \(5.0-\mathrm{g}\) sample of water starting at \(60.0^{\circ} \mathrm{C}\) loses \(418 \mathrm{~J}\) of energy in the form of heat. What is the final temperature of the water after this heat loss? a. \(20 .{ }^{\circ} \mathrm{C}\) b. \(40 .{ }^{\circ} \mathrm{C}\) c. \(50 .{ }^{\circ} \mathrm{C}\) d. \(60 .{ }^{\circ} \mathrm{C}\) e. \(80 .{ }^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The final temperature of the water is \(40.0 \degree \text{C}\), which is option b.

Step 1: Identify the known values

We start with a water sample of mass \(m = 5.0 \, \text{g}\), with an initial temperature \(T_i = 60.0 \degree \text{C}\), and it loses \(q = 418 \, \text{J}\) of heat energy. The specific heat capacity of water is typically \(c = 4.18 \, \text{J/g} \cdot \degree \text{C}.\)

Step 2: Set up the heat transfer equation

The formula for heat transfer is \(q = mc\Delta T\), where \(\Delta T\) is the change in temperature. Since the heat is lost, \(\Delta T = T_f - T_i\) will have a negative value.

Step 3: Solve for the change in temperature, \(\Delta T\)

Rearrange the formula to find \(\Delta T\): \[ \Delta T = \frac{q}{mc} = \frac{-418}{5.0 \times 4.18} \]Calculate \(\Delta T\):\[ \Delta T = \frac{-418}{20.9} = -20 \degree \text{C}\]

Step 4: Determine final temperature \(T_f\)

Using \(\Delta T = -20 \degree \text{C}\), find the final temperature:\[T_f = T_i + \Delta T = 60.0 \degree \text{C} - 20 \degree \text{C} = 40.0 \degree \text{C}\]

Step 5: Select the correct multiple-choice answer

Based on the calculation, the final temperature \(T_f\) is \(40.0 \degree \text{C}\), which corresponds to option b.

Key Concepts

Understanding Specific Heat Capacity

Specific heat capacity is a measure of how much energy in the form of heat is needed to raise the temperature of a given mass of a substance by one degree Celsius. Think of it as the "energy holding capacity" of a material.

For water, this value is quite high, at around \(4.18 \, \text{J/g} \cdot \degree \text{C}\). This means that for every gram of water, you need \(4.18\) joules of energy to heat it by \(1\) Celsius degree. This property makes water an effective substance for tempering changes in temperature. In heating and cooling processes, specific heat capacity helps us quantify and predict how much a substance will warm up or cool down when a certain amount of energy is applied or removed.

• High specific heat means the substance can absorb a lot of heat without a significant change in temperature.
• Compared to metals like iron or aluminum, water's specific heat capacity is higher, which is why oceans can moderate climate by buffering temperature changes.

Understanding specific heat capacity is essential in calculating how a material will behave under thermal stress, which is what we observe in problems like the one we're solving here.

Calculating Temperature Change

When we talk about temperature change, specifically in regard to energy interactions, it's vital to understand the formula that relates energy loss (heat), mass, and specific heat capacity. The equation commonly used is:\[ q = mc\Delta T \]Where:

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

In our example, the water loses heat, implying that \(q\) is negative. Solving for temperature change \((\Delta T)\) involves rearranging the equation to:\[ \Delta T = \frac{q}{mc} \]By plugging in the known values, you can find how much the temperature of the water sample decreases due to the energy loss, demonstrating the direct influence of specific heat capacity on temperature regulation.

Understanding Energy Loss in Heat Transfer

Energy loss in heat transfer concepts is a key part of understanding thermodynamic processes. When heat energy moves out of a system, the energy loss can change the system's temperature. The principles of conservation of energy tell us that this energy must go somewhere, whether into the environment or another system.

• In our example, the water sample loses \(418 \, \text{J}\) of energy, causing its temperature to decrease.
• Heat transfer often occurs from a high-temperature body to a lower-temperature surrounding until thermal equilibrium is reached.

Understanding this balance helps explain not only everyday phenomena, like how an ice cube melts in a drink but also complex systems, such as how cooling systems in engines work. The lost energy is usually transferred via mechanisms like conduction, convection, or radiation. In these processes, specific heat capacity and initial conditions, such as starting temperature and mass of the substance, will determine the final temperature after energy loss occurs. Recognizing this helps anticipate the outcomes of energy exchanges in any thermal process.