/*! 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 106 A tube leads from a \(0.150-\mat... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 106

A tube leads from a \(0.150-\mathrm{kg}\) calorimeter to a flask in which water is boiling under atmospheric pressure. The calorimeter has specific heat capacity 420 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and it originally contains 0.340 \(\mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C}\) . Steam is allowed to condense in the calorimeter at atmospheric pressure until the temperature of the calorimeter and contents reaches \(71.0^{\circ} \mathrm{C}\) , at which point the total mass of the calorimeter and its contents is found to be 0.525 \(\mathrm{kg}\) . Compute the heat of vaporization of water from these data.

Short Answer

Expert verified
The heat of vaporization of water is approximately 2,375,143 J/kg.

Step by step solution

01

Identify Initial and Final Mass

The heat problem is based on the change of mass from steam to water. Initially, the mass in the calorimeter is the sum of the mass of the calorimeter and water: \( m_{i} = 0.150 \, \text{kg} + 0.340 \, \text{kg} = 0.490 \, \text{kg} \). After the steam condenses, the final mass is 0.525 kg. Therefore, the mass of the steam that condensed is \( m_{s} = 0.525 \, \text{kg} - 0.490 \, \text{kg} = 0.035 \, \text{kg} \).
02

Determine Heat Absorbed by Water and Calorimeter

Apply the formula for heat absorbed by the calorimeter and the water originally present: \( Q = (m_{water}c_{water} + m_{cal}c_{cal})(T_{f} - T_{i}) \).Substituting the values: \( Q = \left( 0.340 \times 4186 + 0.150 \times 420 \right)(71.0 - 15.0) \).Calculate to find \( Q = (1422 + 63)(56) = 1485 \times 56 = 83160 \, \text{J} \).
03

Relate Heat with Heat of Vaporization

The heat \( Q \) absorbed by the water and calorimeter equals the heat released by the steam as it condenses, given by \( Q = m_{s}L \), where \( L \) is the heat of vaporization of the water.We have \( 83160 = 0.035 \, L \).Thus, \( L = \frac{83160}{0.035} \approx 2,375,143 \, \text{J/kg} \).

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Key Concepts

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

Calorimetry
Calorimetry is a fascinating area of science that deals with measuring the amount of heat exchanged during physical processes. It鈥檚 like a detective tool for thermal changes.
When you want to know how much heat is absorbed or released, calorimetry is here to help you. Imagine it's a thermometer that not only records temperature but also calculates heat flow. In our exercise, we're looking at a calorimeter setup.
  • The calorimeter is a container 鈥� usually insulated 鈥� that reduces heat transfer with the environment.
  • In the problem, the calorimeter starts with a known mass of water and receives steam from a flask as it reaches equilibrium.
The magic of calorimetry is in its formulas. With the heat equation, we can see how energy moves between steam and water inside the calorimeter. This helps us solve for the heat of vaporization, which is key for converting steam into water through condensation.
Specific Heat Capacity
Specific heat capacity is like a measure of a material's stubbornness in changing temperature! It tells us how much heat is needed to change the temperature of a unit mass by 1 degree Celsius or Kelvin.
In other words, it鈥檚 about how easily something can heat up or cool down.
For this exercise:
  • The specific heat capacity of water is high (4186 J/kg路K). That means water retains and transfers heat effectively, a vital trait in controlling temperature shifts.
  • The calorimeter itself also possesses a specific heat capacity of 420 J/kg路K, calculated separately to understand the heat exchange in myriad parts.
In the context of the calorimetry exercise, knowing these values allows us to find how much the system absorbs until the final temperatures meet. Without these, solving energy problems in calorimetry would feel like chasing wild geese!
Condensation of Steam
Condensation is a process where steam transforms into liquid water, releasing latent heat of vaporization during the change of state. Imagine steam as energetic party-goers with lots of movement, while liquid water is calmer, with less movement.
When steam condenses, it releases a significant amount of this stored energy and slows down.
In our problem:
  • When the steam enters the cooler water in the calorimeter, it condenses, warming the water until equilibrium.
  • At condensation, heat is released, warming the calorimeter and stored water.
The heat exchanged during condensation helps us determine the heat of vaporization of water. This process is vital in calorimetry, as it鈥檚 the bridge that links the steam鈥檚 energetic phase to the liquid, heat-storing phase, offering a neat and practical lesson in energy transformation.

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Most popular questions from this chapter

While running, a \(70-\mathrm{kg}\) student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . To maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , shightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

A capstan is a rotating drum or cylinder over which a rope or cond slides in order to provide a great amplification of the rope's tension while keeping both ends free (Fig. 17.33). Since the added tension in the rope is due to friction, the capstan generates thermal energy. (a) If the difference in tension between the two ends of the rope is 520.0 \(\mathrm{N}\) and the capstan has a diameter of 10.0 \(\mathrm{cm}\) and turns once in 0.900 \(\mathrm{s}\) , find the rate at which thermal energy is generated. Why does the number of turns not matter? (b) If the capstan is made of iron and has mass 6.00 \(\mathrm{kg}\) , at what rate does its temperature rise? Assume that the temperature in the capstan is uniform and that all the thermal energy generated flows into it.

A pot with a steel bottom 8.50 \(\mathrm{mm}\) thick rests on a hot stove.The area of the bottom of the pot is 0.150 \(\mathrm{m}^{2}\) . The water inside the pot is at \(100.0^{\circ} \mathrm{C},\) and 0.390 \(\mathrm{kg}\) are evaporated every 3.00 \(\mathrm{min}\) . Find the temperature of the lower surface of the pot, which is in contact with the stove.

On a cool \(\left(4,0^{\circ} \mathrm{C}\right)\) Saturfay moming, a pilot fills the fuel tanks of her Pitts \(S-2 C\) (a two-seat aerobatic airplane) to their full capacity of 106.0 L. Before flying on Sunday morning, when the temperature is again \(4.0^{\circ} \mathrm{C}\) , she checks the fuel level and finds only 103.4 \(\mathrm{L}\) of gasoline in the tanks. She realizes that it was hot on Saturday afternoon, and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in "C) reached by the fuel and the tank on Saturday aftemoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10^{-4} \mathrm{K}^{-1}\) , and the tank is made of aluminum. (b) In order to have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?

Steam Burns Versus Water Rurns. What is the amount of heat input to your skin when it receives the hear released (a) by 25.0 \(\mathrm{g}\) of steam initially at \(100.0^{\circ} \mathrm{C}\) , when it is cooled to skin temperature \(\left(34.0^{\circ} \mathrm{C}\right) ?\left(\text { b) By } 25.0 \text { g of water initially at } 100.0^{\circ} \mathrm{C}\right.\) when it is cooled to \(34.0^{\circ} \mathrm{C} ?\) (c) What docs this tell you about the relative severity of steam and hot water burns?
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