Problem 77 - College Physics

Question

A thirsty nurse cools a \(2.00 \mathrm{~L}\) bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass \(0.257 \mathrm{~kg}\) and adding \(0.120 \mathrm{~kg}\) of ice initially at \(-15.0^{\circ} \mathrm{C}\). If the soft drink and mug are initially at \(20.0^{\circ} \mathrm{C},\) what is the final temperature of the system. assuming no heat losses?

Step-by-Step Solution

Answer

The final temperature is approximately 1.2°C.

Step 1: Identify System Components and Their Initial States

The system consists of three main components: the soft drink, the aluminum mug, and the ice. The initial temperature of the soft drink and the mug is 20.0°C, while the ice is at -15.0°C. We need to calculate the final equilibrium temperature when these components reach thermal equilibrium.

Step 2: Determine Heat Capacity and Related Constants

- The specific heat capacity of water (soft drink) is approximately 4.18 J/g°C. - The specific heat capacity of aluminum (the mug) is 0.897 J/g°C. - The specific heat capacity of ice is 2.09 J/g°C. - The latent heat of fusion for ice is 334 J/g.

Step 3: Calculate Energy Required to Warm the Ice to 0°C

The mass of the ice is 120 g, and its initial temperature is -15.0°C. The energy required to warm the ice to 0°C is calculated by:\[ Q_1 = m_{ ext{ice}} \times c_{ ext{ice}} \times \Delta T = 120 \times 2.09 \times (0 - (-15)) = 3768 \text{ J}. \]

Step 4: Calculate Energy Required for Ice to Melt

To melt the ice at 0°C, we use the latent heat of fusion:\[ Q_2 = m_{ ext{ice}} \times L_f = 120 \times 334 = 40080 \text{ J}. \]

Step 5: Calculate Energy Changes for the Soft Drink and the Mug

Calculate the heat loss of the soft drink as it cools from 20°C to the final temperature (T):\[ Q_3 = m_{ ext{drink}} \times c_{ ext{drink}} \times (T - 20) = 2000 \times 4.18 \times (T - 20). \]For the aluminum mug from 20°C to T:\[ Q_4 = m_{ ext{mug}} \times c_{ ext{mug}} \times (T - 20) = 257 \times 0.897 \times (T - 20). \]

Step 6: Equate Total Heat Changes to Zero and Solve for T

The total heat changes, considering no heat loss to the environment, should sum to zero:\[ 3768 + 40080 - 2000 \times 4.18 \times (20 - T) - 257 \times 0.897 \times (20 - T) = 0. \]Solve this equation for the final temperature, T.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a key concept when dealing with thermal problems. It describes the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. In this exercise, we deal with three different materials: water (the soft drink), aluminum (the mug), and ice. Each has a different specific heat capacity, reflecting how much heat they each require to change temperatures slightly.

Here's a quick breakdown:

• Water (and similarly, the soft drink) has a specific heat capacity of 4.18 J/g°C.
• Aluminum, a common metal used for its lightweight properties, has a specific heat capacity of 0.897 J/g°C.
• Ice, the solid form of water, has a specific heat capacity of 2.09 J/g°C prior to reaching 0°C and melting.

These values illustrate why different substances heat up or cool down at different rates. The higher the specific heat capacity, the more energy is needed for a temperature change, meaning water cools down slower than aluminum given the same conditions.

Latent Heat of Fusion

Latent heat of fusion is the energy needed to change a substance from solid to liquid at a constant temperature, without changing its temperature. For water, this is a crucial value needed to calculate the energy required for ice to melt. In our scenario, the latent heat of fusion for water is 334 J/g.

This means that for every gram of ice, 334 J of energy are needed to convert it to liquid water at 0°C. This heat does not alter the temperature of the ice; instead, it breaks the molecular bonds that keep the ice in a solid state. In calculations, the energy required for this phase change forms a significant component of the total energy exchanges in the system.

Heat Transfer

Heat transfer is the process of energy moving from a warmer object to a cooler one, until they both reach the same temperature, known as thermal equilibrium. In this exercise, heat transfers occur between the warm soft drink and mug and the cold ice.

Understanding heat transfer involves recognizing:

• Energy flows from higher to lower temperatures.
• In a closed system with no external heat losses, the heat lost by warmer components equals the heat gained by cooler ones.

Calculating this involves summing the energy changes from each component as they reach a common equilibrium temperature. If there's no leak to the environment, every joule lost by the warmer parts of the system goes to warming the cooler parts.

Thermal Energy Calculation

Thermal energy calculation integrates all key concepts discussed, such as specific heat capacity, latent heat of fusion, and heat transfer. This process involves determining the energy changes in the system as it moves towards thermal equilibrium.

Here's how it works in steps:

• Calculate the energy required to heat the ice from its initial temperature to its melting point.
• Determine the energy required to melt the ice using the latent heat of fusion.
• Compute the energy released by the soft drink and the aluminum mug as they cool down to a new equilibrium temperature.

You then set up an equation where the total energy gained by the ice equals the total energy lost by the soft drink and the mug, ensuring the sum of heat changes in the system is zero. Solving this provides the final equilibrium temperature of the system.