/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 An ice chest at a beach party co... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The final temperature is approximately \(18.36^{\circ}\mathrm{C}\).

Step by step solution

01

Understanding the Problem

The problem is about finding the final equilibrium temperature when two substances, soda and a watermelon, at different temperatures are mixed in an insulated environment. We need to use the concept of conservation of energy, assuming no heat loss to the surroundings.
02

Define the Initial Parameters

1. Number of soda cans: 122. Mass of each soda can: \(0.35\, \mathrm{kg}\)3. Total mass of soda: \(m_s = 12 \times 0.35 = 4.2 \, \mathrm{kg}\)4. Initial temperature of soda: \(T_s = 5.0^{\circ}\mathrm{C}\)5. Mass of watermelon: \(m_w = 6.5 \, \mathrm{kg}\)6. Initial temperature of watermelon: \(T_w = 27^{\circ}\mathrm{C}\)7. Specific heat capacity of soda and watermelon: \(c = 3800 \, \mathrm{J}/(\mathrm{kg} \cdot \mathrm{C}^{\circ})\)
03

Set up Energy Balance Equation

Since no heat is lost to the surroundings, the heat lost by the watermelon equals the heat gained by the sodas:\[m_w \cdot c \cdot (T_f - T_w) = - m_s \cdot c \cdot (T_f - T_s)\]Where:- \( T_f \) is the final equilibrium temperature- \( T_w \) and \( T_s \) are the initial temperatures of the watermelon and soda, respectively.
04

Solve for the Final Temperature

Substitute the known values into the equation:\[6.5 \times 3800 \times (T_f - 27) = - 4.2 \times 3800 \times (T_f - 5)\]Simplify the equation by canceling \(3800\):\[6.5 \cdot (T_f - 27) = - 4.2 \cdot (T_f - 5)\]Distribute and solve for \(T_f\):\[6.5T_f - 175.5 = -4.2T_f + 21\]Add \(4.2T_f\) to both sides:\[6.5T_f + 4.2T_f = 175.5 + 21\]Which simplifies to:\[10.7T_f = 196.5\]Divide both sides by \(10.7\):\[T_f = \frac{196.5}{10.7} \approx 18.36^{\circ}\mathrm{C}\]
05

Conclude the Solution

The final equilibrium temperature of the ice chest, where both the watermelon and soda are at the same temperature, is approximately \(18.36^{\circ}\mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a property of a material that indicates how much heat energy is required to change the temperature of a certain amount of that material by one degree. In simple terms, it's a measure of a material's ability to absorb and store heat. The specific heat capacity is usually expressed in Joules per kilogram per degree Celsius (J/kg°C). For example, in the exercise, both the soda and the watermelon have a specific heat capacity of 3800 J/(kg°C), meaning that 3800 joules of energy are needed to change 1 kg of either substance by 1°C.

When solving problems like the one in the exercise, knowing the specific heat capacity allows us to calculate how much heat is exchanged when two bodies at different temperatures come into contact. The rule of thumb is that substances with higher specific heat capacities require more energy to change their temperatures, compared to those with lower capacities.
  • High specific heat capacity: Stores more heat energy, resists temperature change.
  • Low specific heat capacity: Stores less heat energy, changes temperature easily.
Understanding specific heat capacity is essential for predicting how different materials will react when subjected to thermal conditions, making it an important concept in thermodynamics.
Conservation of Energy
The principle of conservation of energy is fundamental in physics and thermodynamics. It states that energy cannot be created or destroyed, only transformed from one form to another. This principle is applied in many fields, including when solving the exercise involving the soda and the watermelon.

In this exercise, the conservation of energy tells us that the total energy within the isolated system (soda and watermelon in an ice chest) remains constant. When the warm watermelon comes into contact with the cooler sodas, energy is transferred in the form of heat. The watermelon loses heat, while the sodas gain it, until thermal equilibrium is reached.
The equation used in our step-by-step solution demonstrates this balance:\[m_w \cdot c \cdot (T_f - T_w) = - m_s \cdot c \cdot (T_f - T_s)\]
This equation confirms that the energy lost by the watermelon equals the energy gained by the sodas. It's crucial to ensure full energy balance because any discrepancies would indicate an error, perhaps due to external heat loss or measurement inaccuracies.
  • No net loss or gain in total energy within a closed system.
  • Ensures accurate energy distribution calculations.
By relying on energy conservation, we predict the eventual temperature the system will stabilize at—also known as equilibrium temperature.
Equilibrium Temperature
Equilibrium temperature is the final uniform temperature reached by two or more materials when they come into contact and exchange heat. It occurs when there is no net heat flow between the bodies, meaning the temperatures are equal. In our exercise, this is the temperature at which both the watermelon and the soda cans stabilize after being in the ice chest together.

Achieving equilibrium temperature involves understanding the interactions between the heat capacities, masses, and initial temperatures of the involved substances. As seen in the exercise, you can find the equilibrium temperature through an energy balance equation that considers all these factors.
  • The net heat exchange continues until equilibrium is reached.
  • The equilibrium temperature depends on initial conditions and material properties.
The equation used in the solution is crucial for determining this temperature, as it equates the heat lost by one body to the heat gained by another. After solving it, we find that the equilibrium temperature for this system is approximately 18.36°C. This solution gives insight into how thermal systems balance themselves out over time.

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Most popular questions from this chapter

A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 240 Calories \((1\) Calorie \(=4186 \mathrm{J})\). What mass of perspiration would one have to lose to get rid of this energy? At body temperature, the latent heat of vaporization of water is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\).

The heating element of a water heater in an apartment building has a maximum power output of \(28 \mathrm{kW}\). Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of \(14 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\). If the water going into the heater has a temperature of \(11{ }^{\circ} \mathrm{C},\) what is the maximum possible temperature of the hot water that each showering resident receives?

Two bars of identical mass are at \(25^{\circ} \mathrm{C} .\) One is made from glass and the other from another substance. The specific heat capacity of glass is \(840 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) When identical amounts of heat are supplied to each, the glass bar reaches a temperature of \(88{ }^{\circ} \mathrm{C},\) while the other bar reaches \(250.0^{\circ} \mathrm{C} .\) What is the specific heat capacity of the other substance?

When the temperature of a coin is raised by \(75 \mathrm{C}^{\circ},\) the coin's diameter increases by \(2.3 \times 10^{-5} \mathrm{m} .\) If the original diameter of the coin is \(1.8 \times 10^{-2} \mathrm{m},\) find the coefficient of linear expansion.

The latent heat of vaporization of \(\mathrm{H}_{2} \mathrm{O}\) at body temperature \(\left(37.0^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) To cool the body of a \(75-\mathrm{kg}\) jogger [average specific heat capacity \(\left.=3500 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\right]\) by \(1.5 \mathrm{C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

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