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Chapter 5: Problem 22

In a thermodynamic study, a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? (b) If the inlet and outlet were closed, what type of system would it be?

Short Answer

Expert verified
(a) The solution in the apparatus is an \(open\) system since it can exchange both matter and energy with its surroundings. (b) If the inlet and outlet were closed, the system would become a \(closed\) system as it can still potentially exchange energy with its surroundings, but not matter.

Step by step solution

01

Definition of Closed, Open, and Isolated Systems

In thermodynamics, there are three types of systems based on the exchange of matter and energy with their surroundings: 1. Closed system: A system that can exchange energy, but not matter, with its surroundings. 2. Open system: A system that can exchange both matter and energy with its surroundings. 3. Isolated system: A system that cannot exchange either matter or energy with its surroundings.
02

Identify the Type of System in the Apparatus (a)

In the given system, a solution is continuously flowing into the apparatus at the top and out at the bottom. This implies that there is an exchange of matter happening between the system and its surroundings. Since matter is being exchanged, it cannot be a closed or isolated system. Therefore, the solution in the apparatus is an open system.
03

Identify the Type of System If Inlet and Outlet Were Closed (b)

If the inlet and outlet were closed, the solution would no longer flow into or out of the apparatus. This means that the exchange of matter between the system and its surroundings is stopped. However, there is no information that suggests that the exchange of energy is stopped as well. So, if the inlet and outlet were closed, the system would become a closed system as it can still potentially exchange energy with its surroundings.

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Key Concepts

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

Open System
In thermodynamics, an open system is a type that can exchange both matter and energy with its surroundings. Imagine a pot of water on a stove, with the lid off. The water can evaporate into the air, and heat from the stove moves into the pot. This demonstrates how both matter (water vapor) and energy (heat) can move in and out of the system.
  • **Matter Exchange:** The ongoing exchange of substances, like how the solution in the exercise flows in and out of the apparatus.
  • **Energy Exchange:** It can absorb or release heat or work with its environment.

Open systems are common in daily life since many systems exchange both matter and energy to perform processes.
Closed System
A closed system in thermodynamics is able to exchange energy but not matter with its surroundings. Picture a sealed pot on a stove, this setup allows heat to pass through but prevents the contents from escaping. This means the pot can change temperature as it gains or loses heat, even though the mass inside remains constant.
  • **No Matter Exchange:** The sealed environment keeps all substances inside, just like when you close the inlet and outlet of the apparatus in the original exercise.
  • **Energy Exchange:** Heat or work can cross the boundary even though no physical material can.

Closed systems often appear in experiments when isolating specific reactions or processes, ensuring mass stays constant while energy changes occur.
Isolated System
An isolated system does not exchange either matter or energy with its surroundings. It is completely self-contained. Think of a thermos flask filled with hot soup. Ideally, the soup remains hot for a long time as it's neither losing heat to the surroundings nor gaining heat from them, and no matter leaks in or out.
  • **No Matter Exchange:** The boundary is perfectly sealed against material movement.
  • **No Energy Exchange:** Heat or work cannot penetrate, making it insulated to energy transfers.

Although true isolated systems are theoretical鈥攕ince absolute isolation is difficult to achieve in practice鈥攖hey provide a useful model for understanding thermodynamic laws.

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Most popular questions from this chapter

(a) According to the first law of thermodynamics, what quantity is conserved? (b) What is meant by the internal energy of a system? (c) By what means can the internal energy of a closed system increase?

A magnesium ion, Mg \(^{2+},\) with a charge of \(3.2 \times 10^{-19} \mathrm{Cand}\) an oxide ion, \(\mathrm{O}^{2-},\) with a charge of \(-3.2 \times 10^{-19} \mathrm{C}\) , are separated by a distance of 0.35 \(\mathrm{nm}\) . How much work would be required to increase the separation of the two ions to an infinite distance?

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) 1 \(\mathrm{mol} \mathrm{CO}_{2}(s)\) or 1 \(\mathrm{mol} \mathrm{CO}_{2}(g)\) at the same temperature, ( b) 2 \(\mathrm{mol}\) of hydrogen atoms or 1 \(\mathrm{mol}\) of \(\mathrm{H}_{2},(\mathbf{c}) 1 \mathrm{mol} \mathrm{H}_{2}(g)\) and 0.5 \(\mathrm{mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{mol} \mathrm{N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{N}_{2}(g)\) at \(300^{\circ} \mathrm{C}\) .

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at \(25.1^{\circ} \mathrm{C} .\) A \(121.0-\mathrm{g}\) block of copper metal is heated to \(100.4^{\circ} \mathrm{C}\) by putting it in a beaker of boiling water. The specific heat of \(\mathrm{Cu}(s)\) is \(0.385 \mathrm{J} / \mathrm{g}-\mathrm{K}\) . The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of \(30.1^{\circ} \mathrm{C}\) (a) Determine the amount of heat, in J, lost by the copper block. (b) Determine the amount of heat gained by the water. The specific heat of water is \(4.18 \mathrm{J} / \mathrm{g}-\mathrm{K}\) . (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K. (d)What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

Under constant-volume conditions, the heat of combustion of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is 15.57 \(\mathrm{kJ} / \mathrm{g}\) . A 3.500 -g sample of glucose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to \(24.72^{\circ} \mathrm{C}\) (a) What is the total heat capacity of the calorimeter? (b) If the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?
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