Problem 42 - Physics

Question

An ice chest at a beach party contains 12 cans of soda at \(5.0^{\circ} \mathrm{C}\). Each can of soda has a mass of \(0.35 \mathrm{~kg}\) and a specific heat capacity of \(3800 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). Someone adds a \(6.5-\mathrm{kg}\) watermelon at \(27{ }^{\circ} \mathrm{C}\) to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature \(T\) of the soda and watermelon.

Step-by-Step Solution

Answer

The final temperature of the soda and watermelon is approximately 18.9°C.

Step 1: Define the Problem

We need to find the final equilibrium temperature \(T\) when a watermelon is added to the ice chest containing soda. We assume that the heat lost by the watermelon is equal to the heat gained by the soda, as there is no heat exchange with the surroundings.

Step 2: Set Up the Heat Balance Equation

Apply the principle of conservation of energy: the total heat absorbed by the soda equals the total heat lost by the watermelon. Define \(Q_{soda}\) and \(Q_{watermelon}\) as follows:\[Q_{soda} = m_{soda} \times c_{soda} \times (T - T_{initial, soda})\]\[Q_{watermelon} = m_{watermelon} \times c_{watermelon} \times (T_{initial, watermelon} - T)\]Since \(Q_{soda} = Q_{watermelon}\), we can equate these expressions.

Step 3: Calculate Mass and Initial Parameters

There are 12 soda cans, each with a mass of \(0.35 \text{ kg}\), so the total mass \(m_{soda} = 12 \times 0.35 = 4.2 \text{ kg}\). The specific heat capacity is \(c_{soda} = 3800 \text{ J/(kg}\cdot \text{°C)}\). The initial temperature \(T_{initial, soda} = 5.0 \text{ °C}\).For the watermelon, \(m_{watermelon} = 6.5 \text{ kg}\), \(c_{watermelon} = 4200 \text{ J/(kg}\cdot \text{°C)}\), and \(T_{initial, watermelon} = 27.0 \text{ °C}\).

Step 4: Substitute and Solve the Equation

Plug in the known values to the equation from Step 2, and solve for \(T\):\[4.2 \times 3800 \times (T - 5) = 6.5 \times 4200 \times (27 - T)\]Solving this equation will give us the final temperature \(T\).

Step 5: Simplify and Calculate

Simplify and solve the equation for \(T\):\[15960(T - 5) = 27300(27 - T)\]Expanding both sides gives:\[15960T - 79800 = 738600 - 27300T\]Combine like terms:\[43260T = 818400\]Solve for \(T\):\[T = \frac{818400}{43260} \approx 18.9 \text{ °C}\]

Step 6: Verify the Solution

Ensure that the calculated temperature is reasonable considering the context. The final temperature \(18.9 \text{ °C}\) is between 5°C and 27°C, which makes sense given the heat exchange between the soda and the watermelon.

Key Concepts

Specific Heat Capacity

The specific heat capacity is an important concept in thermodynamics that defines how much heat energy is required to change the temperature of a substance by one degree Celsius. In our problem, to calculate the heat exchange, we use the specific heat capacities of soda and watermelon.
The soda has a specific heat capacity of 3800 J/(kg·°C), which tells us how much energy in joules is needed to increase the temperature of 1 kg of soda by 1°C. In the case of the watermelon, its specific heat capacity is similar to water, which is around 4200 J/(kg·°C). This means the watermelon can hold a considerable amount of heat before its temperature changes.
These values are essential for determining how substances within the ice chest interact in terms of energy transfer. The specific heat capacity is greatly affected by the type of material, and this value allows us to predict how different materials will behave in thermal processes.

Heat Transfer

Heat transfer is the movement of thermal energy from one body or system to another. In our scenario, it's the transfer of thermal energy between the soda and the watermelon.
There are three main modes of heat transfer: conduction, convection, and radiation. However, in this closed system example, the primary concern is conduction, as it involves direct contact transfer of heat between the soda cans and the watermelon.
Due to the difference in initial temperatures, the watermelon, being warmer, will lose heat, while the soda, being cooler, will gain heat. This transfer continues until thermal equilibrium is reached, ensuring the energy lost by the watermelon equals the energy gained by the soda. Understanding heat transfer helps explain why their final temperature based on these interactions lies between their initial temperatures.

Conservation of Energy

The law of conservation of energy is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed, only transformed from one form to another or transferred between objects.
In our exercise, the total energy within the chest remains constant, as mentioned. When the watermelon is added, the energy from its warmth is transferred to the soda, aligning perfectly with the conservation of energy. No energy leaves or enters the system, meaning that any heat loss by the warm watermelon must equal the heat gained by the cool soda.
This principle forms the basis for setting up the heat balance equations in our problem, ensuring that the calculations predict the final state accurately, reinforcing the reliability of solutions derived from energy balance methods.

Equilibrium Temperature

Equilibrium temperature is the final temperature achieved when two substances reach a state where no heat flows between them. In our context, it's the temperature at which the soda and watermelon stabilize.
When using the related formulas, the equilibrium temperature is determined by balancing the heat lost and gained until the rate of heat coming into one equals the heat leaving the other. In this case, it resolves to around 18.9 °C.
This concept illustrates how systems naturally progress towards equilibrium, highlighting the universal drive for balance in energy distribution across a system. It's essential for understanding how mixtures or combined substances respond to thermal interactions, demonstrating predictable outcomes of energy exchange processes.