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In a container of negligible mass, 0.0400 kg of steam at 100\(^\circ\)C and atmospheric pressure is added to 0.200 kg of water at 50.0\(^\circ\)C. (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

Short Answer

Expert verified
The final temperature is 100°C. There are 0.240 kg of liquid water and no steam.

Step by step solution

01

Write the Heat Exchange Equation

When the steam condenses to water, it releases heat, and this heat is absorbed by the cooler water, possibly reaching a common equilibrium temperature. Denote the final temperature as \( T_f \). Use:\[ Q_{released} = Q_{absorbed} \]
02

Calculate Heat Released by Steam

Steam initially releases heat when it condenses to water. Use the latent heat of vaporization: \[ Q_{condensation} = m_{steam} \cdot L_v \]where \( m_{steam} = 0.0400 \text{ kg} \) and \( L_v = 2260 \text{ kJ/kg} \) for water.
03

Calculate Heat Released When Cooled to Final Temperature

After condensation, the temperature of the new water (from steam) is lowered to \( T_f \). Use:\[ Q_{cooling} = m_{water} \cdot c \cdot (T_{initial} - T_f) \]where \( c = 4.18 \text{ kJ/kg} \cdot \text{°C} \) and \( T_{initial} = 100\, ^{\circ}\text{C} \).
04

Calculate Heat Absorbed by Initial Water

The initial water temperature is increased from 50.0 \( ^{\circ} \text{C} \) to \( T_f \). \[ Q_{absorbed} = m_{water} \cdot c \cdot (T_f - T_{initial}) \]where \( m_{water} = 0.200 \text{ kg} \) and \( T_{initial} = 50.0 \text{°C} \).
05

Equate Heat Released and Absorbed

Set the heat released by the steam equal to the heat absorbed by the initial water:\[ Q_{condensation} + Q_{cooling} = Q_{absorbed} \]Solve this equation to find \( T_f \).
06

Determine the Final Quantity of Steam and Water

Since the steam condenses to water, there will be more liquid water and no remaining steam. The total mass of liquid water is the sum of both initial and condensed parts.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics. It describes the movement of thermal energy from one object or substance to another due to a temperature difference between them. In the given exercise, heat transfer occurs when steam at 100°C is added to water at 50°C.

- **Direction of Heat Flow**: Heat naturally flows from a hotter object to a cooler one. Here, this means that the steam (hotter) will transfer heat to the cooler water.
- **Mechanism**: Heat can be transferred through different methods such as conduction, convection, and radiation. But in this exercise, conduction is the main method because heat is transferred directly between the steam and the water.
- **No Loss Assumption**: It’s assumed that no heat is lost to the surroundings. This is an idealized situation meaning all the heat stays within the system and none escapes into the air or container.
Latent Heat
Latent heat is the amount of heat required to change the state of a substance without changing its temperature. This concept is evident when the steam condenses into water.

- **Latent Heat of Vaporization**: For water, the latent heat of vaporization is crucial. It’s the heat necessary to convert water in gaseous state (steam) to liquid.
- **Calculation**: The latent heat of vaporization for water is 2260 kJ/kg. In this exercise, it tells us how much energy is released when steam changes into liquid water.
- **Energy Release**: When steam condenses to water, it releases this latent energy, which is then used to heat the remaining water solution.
Phase Change
A phase change occurs when a substance changes from one state of matter to another, such as from gas to liquid. In this exercise, the steam undergoes a phase change as it condenses into water.

- **From Gas to Liquid**: The initial state of steam is gas. When heat is released during condensation, the steam transforms into liquid water.
- **Impact on Temperature**: During this change, the temperature of the steam doesn’t fall until it has turned completely into water. This is because latent heat is involved, maintaining the 100°C temperature until completion of the phase shift.
- **Final Outcome**: After the condensation process completes, the new liquid formed by the condensed steam mixes with the original water, affecting the overall temperature of the system.
Equilibrium Temperature
Equilibrium temperature is the uniform temperature a system reaches when no more net heat transfer occurs within the system. In this exercise, the system reaches a common equilibrium temperature between the condensed steam and initial water.

- **Reaching Equilibrium**: As heat is exchanged between the steam and the water, their initial temperature differences reduce until they balance out.
- **Calculation**: Using the formula: \( Q_{released} = Q_{absorbed} \), where the energy released by the steam is set equal to the energy absorbed by the water, an equilibrium temperature is found.
- **Final Measurement**: The equilibrium temperature tells us how warm the total amount of water becomes after all energy exchanges and the phase change completes. This is achieved without loss of heat to the environment as per the problem's conditions.

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Most popular questions from this chapter

Consider a poor lost soul walking at 5 km/h on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 W, and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k'A{_s}{_k}{_i}{_n}(T{_a}{_i}{_r} - T{_s}{_k}{_i}{_n})\), where \(k'\) is 54 J/h \(\cdot\) C\(^\circ\) \(\cdot\) m\(^2\), the exposed skin area \(A{_s}{_k}{_i}{_n}\) is 1.5 m\(^2\), the air temperature \(T{_a}{_i}{_r} \)is 47\(^\circ\)C, and the skin temperature \(T{_s}{_k}{_i}{_n}\) is 36\(^\circ\)C; (iii) the skin absorbs radiant energy from the sun at a rate of 1400 W/m\(^2\); (iv) the skin absorbs radiant energy from the environment, which has temperature 47\(^\circ\)C. (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e\) = 1 and that the skin temperature is initially 36\(^\circ\)C. Which mechanism is the most important? (b) At what rate (in L/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at 36\(^\circ\)C is \(2.42 \times 10{^6}\) J/kg.) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 m\(^2\). What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

If the air temperature is the same as the temperature of your skin (about 30\(^\circ\)C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70-kg person's body produces energy at a rate of about 500 W due to metabolism, 80% of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750-mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 kg.)

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

A spherical pot contains 0.75 L of hot coffee (essentially water) at an initial temperature of 95\(^\circ\)C. The pot has an emissivity of 0.60, and the surroundings are at 20.0\(^\circ\)C. Calculate the coffee’s rate of heat loss by radiation.

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