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Chapter 17: Problem 52

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?

Short Answer

Expert verified
Steam burns are more severe as they release more heat due to phase change.

Step by step solution

01

Understanding the Concept

The problem involves calculating the heat released by steam and water separately when they are cooled. The heat released when steam condenses and the heat consumed to cool down the resultant water have distinct processes: 1) release of latent heat during condensation, 2) cooling of water from condensation, and 3) cooling of water from original water. We will use these steps to find the total heat for steam and water.
02

Calculating Heat Released by Steam Condensation

To find the heat released by 25.0 g of steam as it condenses, use the formula: \( Q = mL \), where \( m = 25.0 \) g, and \( L = 2260 \) J/g, the latent heat of vaporization. So, \( Q = 25.0 \times 2260 = 56500 \) J.
03

Cooling the Condensed Water to Skin Temperature

The steam that condensed into water is now cooled from 100.0掳C to 34.0掳C. Use the formula \( Q = mc\Delta T \), where \( m = 25.0 \) g, \( c = 4.18 \) J/g掳C, and \( \Delta T = 100.0掳C - 34.0掳C = 66.0掳C \). Therefore, \( Q = 25.0 \times 4.18 \times 66.0 = 6897 \) J.
04

Total Heat Released by Steam

Add the heat released during condensation to the heat released when the condensed water is cooled: \( 56500 \text{ J} + 6897 \text{ J} = 63397 \text{ J} \).
05

Calculate Heat Released by Water

For water at 100.0掳C cooling to 34.0掳C, use the formula \( Q = mc\Delta T \), where \( m = 25.0 \) g, \( c = 4.18 \) J/g掳C, and \( \Delta T = 66.0 \)掳C. Therefore, \( Q = 25.0 \times 4.18 \times 66.0 = 6897 \) J.
06

Comparison and Conclusion

Compare the heat absorbed by the skin in both cases: Steam releases \( 63397 \text{ J} \) while water releases \( 6897 \text{ J} \). The amount of heat from steam is significantly greater due to the additional heat from the phase transition from steam to water.

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Key Concepts

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

Phase Transitions
A phase transition occurs when a substance changes from one state of matter to another, like solid to liquid or liquid to gas. In this exercise, we're dealing with the transition from steam (gas) to water (liquid).
The key element in phase transitions is that they involve the absorption or release of energy, called latent heat.
When steam condenses into water, it releases a large amount of energy, known as latent heat of vaporization. This phase transition from steam to water significantly increases the amount of heat transferred to the skin.
  • Steam to water transition releases energy.
  • This process involves latent heat.
The immense energy release during phase transitions makes steam burns more severe than burns from hot water.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one object to another and is a critical concept in understanding burns.
In our problem, heat is transferred from steam or hot water to the skin, causing potential burns.
There are three modes of heat transfer: conduction, convection, and radiation. However, here, we focus on the conduction occurring when steam or water comes in contact with the skin.
Heat transfer is calculated using the formula: \[ Q = mc\Delta T \]where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
  • Heat flows from the hotter steam or water to cooler skin.
  • Heat transfer can cause damage like burns.
Understanding how heat is transferred can help in explaining why steam burns cause more damage.
Latent Heat
Latent heat is the heat absorbed or released during a phase transition without changing the temperature.
In the context of steam, the latent heat of vaporization is a key factor. When steam condenses to water, it releases latent heat, which is a large amount of energy.
The formula to calculate this energy release is: \[ Q = mL \]where \( Q \) is the amount of heat, \( m \) is the mass, and \( L \) is the latent heat of vaporization.
  • The latent heat of vaporization for water is very high at \( 2260 \text{ J/g} \).
  • This additional heat from phase change is what makes steam burns more severe.
Understanding latent heat is crucial to explaining why a small amount of steam can cause significant burns.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius.
For water, the specific heat capacity is \( 4.18 \text{ J/g掳C} \).
In our scenario, once steam condenses to water, the resultant water further cools, releasing more heat during this process as per the formula:\[ Q = mc\Delta T \]where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.
  • The specific heat explains how much energy is needed to cool down the water after steam condensation.
  • This process adds to the total heat imparted to the skin, making burns worse.
Specific heat capacity helps understand the energy transferred during cooling after a phase transition.

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Most popular questions from this chapter

Spacecraft Reentry. A spacecraft made of aluminum circles the earth at a speed of 7700 \(\mathrm{m} / \mathrm{s}\) . (a) Find the ratio of its kinetic energy to the energy required to raise its temperature from \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\) . (The melting point of aluminum is \(660^{\circ} \mathrm{C}\) . Assume a constant specific heat of \(910 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) (b) Discuss the bearing of your answer on the problem of the reentry of a manned space vehicle into the earth's atmosphere.

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200-\mathrm{W}\) electric immersion heater in 0.320 \(\mathrm{kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C}\) ? (b) How much time is required? Assume that all of the heater's power goes into heating the water.

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 metal rod that is 30.0 \(\mathrm{cm}\) long expands by 0.0650 \(\mathrm{cm}\) when its temperature is raised from \(0.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\) . A rod of a different metal and of the same length expands by 0.0350 \(\mathrm{cm}\) for the same rise in temperature. A third rod, also 30.0 \(\mathrm{cm}\) long, is made up of pieces of each of the above metals placed end to end and expands 0.0580 \(\mathrm{cm}\) between \(0.0^{\circ} \mathrm{C}\) and \(100.0^{\circ} \mathrm{C}\) . Find the length of each portion of the composite rod.

A refrigerator door is opened and room-temperature air \(\left(20.0^{\circ} \mathrm{C}\right)\) fills the \(1.50-\mathrm{m}^{3}\) compartment. A \(10.0-\mathrm{kg}\) turkey, also at room temperature, is placed in the refrigerator and the door is closed. The density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . Assume the specific heat of a turkey, like that of a human, is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . How much heat must the refrigerator remove from its compartment to bring the air and the turkey to thermal equilibrium at a temperature of \(5.00^{\circ} \mathrm{C}\) ? Assume no heat exchange with the surrounding environment.
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