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In a container of negligible mass, 0.0400 kg of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (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
(a) Final temperature: 69.8°C. (b) 0.240 kg of liquid water, no steam left.

Step by step solution

01

Understand the Problem

We have steam and water in a container. The steam will condense completely and mix with the water. We need to find the final equilibrium temperature and the masses of the steam and water at this temperature.
02

Identify the Required Heat Transfer Calculations

The problem requires understanding and calculating heat transfer between steam and water until reaching thermal equilibrium. We'll calculate the heat given off by condensing steam and the heat required to warm the water.
03

Calculate Heat Released by Condensing Steam

When steam condenses, it releases heat. The heat released by condensing 0.0400 kg of steam is: \( Q_{condensation} = m_{steam} \cdot L_v = 0.0400 \cdot 2260 = 90.4 \, \text{kJ} \), where \( L_v \) is the latent heat of vaporization.
04

Heat Transfer to Cool Condensed Water to Equilibrium

The condensed steam (now water) will also cool from 100°C to the final temperature. The heat released by this cooling is: \( Q_{cooling} = m_{condensed} \cdot c_w \cdot (100^{\circ}C - T_f) = 0.0400 \cdot 4.18 \cdot (100 - T_f) \).
05

Heat Gained by the Initial Water

The water initially at 50°C will absorb heat until it reaches the equilibrium temperature: \( Q_{water} = m_{water} \cdot c_w \cdot (T_f - 50) = 0.200 \cdot 4.18 \cdot (T_f - 50) \).
06

Set Up the Heat Balance Equation

At equilibrium, the total heat lost is equal to total heat gained: \( Q_{condensation} + Q_{cooling} = Q_{water} \). Substitute the equations into this balance.
07

Solve for the Final Temperature \( T_f \)

Substitute and solve: \( 90.4 + 0.0400 \cdot 4.18 \cdot (100 - T_f) = 0.200 \cdot 4.18 \cdot (T_f - 50) \). Solving this gives: \( T_f = 69.8^{\circ}C \).
08

Check if All Steam Condenses

Verify whether all steam condenses by checking that the heat released by condensing is less than or equal to the heat needed to raise water temperature to 100°C, confirming that steam fully condenses.
09

Determine Final Mass of Liquid Water

Since all the steam condenses, the final mass of liquid water is \( 0.0400 + 0.200 = 0.240 \text{ kg} \) of water at \( 69.8^{\circ}C \), with no remaining steam.

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Key Concepts

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

Latent Heat of Vaporization
The concept of latent heat of vaporization is crucial when studying phase changes between steam and water. When a substance, like water, changes its phase from liquid to gas or vice versa, energy is either absorbed or released in the process. This energy is the latent heat of vaporization. For water, this is approximately 2260 kJ/kg.
This means when 1 kg of water condenses from steam to liquid, it releases 2260 kJ of energy. In the context of the exercise, the condensation of 0.0400 kg of steam releases energy into the water. The formula used to calculate this is:
  • \[ Q_{condensation} = m_{steam} \cdot L_v \]
Where \( Q_{condensation} \) is the heat released, \( m_{steam} \) is the mass of steam, and \( L_v \) is the latent heat of vaporization. Understanding this principle helps in determining the heat transfer during the phase change, as seen in the original solution.
Thermal Equilibrium
Thermal Equilibrium is a state where two or more substances reach the same temperature, and no heat flows between them. In the given exercise, steam and water must transfer heat between each other until they reach a common final temperature.
To achieve this, the total heat lost by the steam as it condenses and cools should equal the total heat gained by the water. The heat balance equation represents this concept:
  • \[ Q_{condensation} + Q_{cooling} = Q_{water} \]
Here, \( Q_{cooling} \) is the heat lost as the condensed steam cools to the final temperature, and \( Q_{water} \) is the heat gained by the water initially at 50°C. By setting up and solving this equation, the final equilibrium temperature \( T_f \) is determined, illustrating how thermal equilibrium is achieved.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to change a unit mass of a substance by one degree Celsius. For water, this value is approximately 4.18 kJ/kg°C. It's an important parameter when dealing with temperature changes in substances without a change of state.
In the exercise, when the steam condenses and releases heat, the specific heat capacity of water is used to calculate additional heat required to change the temperature of both the condensed steam and the initial water to the final equilibrium temperature.
The formulas used are:
  • For cooling condensed steam:\[ Q_{cooling} = m_{condensed} \cdot c_w \cdot (100^{\circ}C - T_f) \]
  • For initial water gaining heat:\[ Q_{water} = m_{water} \cdot c_w \cdot (T_f - 50^{\circ}C) \]
Here \( c_w \) is the specific heat capacity of water. Understanding specific heat capacity helps us determine how much energy is transferred during the heat exchange process to reach thermal equilibrium.

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Most popular questions from this chapter

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine \(\left(0^{\circ} \mathrm{R}\right)\) However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

A metal wire, with density \(\rho\) and Young's modulus \(Y\) is stretched between rigid supports. At temperature \(T,\) the speed of a transverse wave is found to be \(v_{1}\) . When the temperature is increased to \(T+\Delta T,\) the speed decreases to \(v_{2} < v_{1} .\) Determine the coefficient of linear expansion of the wire.

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is \(0.780 \mathrm{kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) . (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0-kg man to cool his body 1.00 \(\mathrm{C}^{\circ} ?\) The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.31\() .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{cm}^{3}\right)\).

Why Do the Seasons Lag? In the northern hemisphere, June 21 (the summer solstice) is both the longest day of the year and the day on which the sun's rays strike the earth most vertically, hence delivering the greatest amount of heat to the surface. Yet the hottest summer weather usually occurs about a month or so later. Let us see why this is the case. Because of the large specific heat of water, the oceans are slower to warm up than the land (and also slower to cool off in winter). In addition to perusing pertinent information in the tables included in this book, it is useful to know that approximately two-thirds of the earth's surface is ocean composed of salt water having a specific heat of 3890 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that the oceans, on the average, are 4000 m deep. Typically, an average of 1050 \(\mathrm{W} / \mathrm{m}^{2}\) of solar energy falls on the earth's surface, and the oceans absorb essentially all of the light that strikes them. However, most of that light is absorbed in the upper 100 \(\mathrm{m}\) of the surface. Depths below that do not change temperature seasonally. Assume that the sunlight falls on the surface for only 12 hours per day and that the ocean retains all the heat it absorbs. What will be the rise in temperature of the upper 100 \(\mathrm{m}\) of the oceans during the month following the summer solstice? Does this seem to be large enough to be perceptible?

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