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Chapter 17: Problem 52

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

Short Answer

Expert verified
Steam transfers 63407.5 J, while water transfers 6907.5 J. Steam burns are more severe.

Step by step solution

01

Understanding the Problem

We need to calculate the amount of heat released by 25.0 g of steam and 25.0 g of water as they both cool down to 34.0掳C. For steam, we must account for both the heat released as it condenses and cools. For water, we only consider cooling.
02

Calculate Heat Released by Steam Condensation and Cooling

First, calculate the heat released by the phase change and cooling of steam. The total heat will be the sum of the heat of condensation and the cooling of the resulting water.- **Condensation:** The heat released by the steam as it condenses into water is calculated using the formula \( q_\text{condensation} = m \cdot L_v \), where \( m \) is mass and \( L_v = 2260 \text{ J/g} \) is the latent heat of vaporization.- **Cooling of Water:** The heat released by the cooling of water from 100掳C to 34掳C is given by \( q_\text{water cooling} = m \cdot c \cdot \Delta T \), where \( c = 4.186 \text{ J/(g路掳C)} \) is the specific heat and \( \Delta T = 66 \text{ 掳C} \).Calculation:- \( q_\text{condensation} = 25.0 \text{ g} \times 2260 \text{ J/g} = 56500 \text{ J} \)- \( q_\text{water cooling} = 25.0 \text{ g} \times 4.186 \text{ J/(g路掳C)} \times 66 \text{掳C} = 6907.5 \text{ J} \)- **Total Heat from Steam:** \( 56500 \text{ J} + 6907.5 \text{ J} = 63407.5 \text{ J} \)
03

Calculate Heat Released by Water Cooling

For water, only the cooling needs to be considered, as there is no phase change. Use the formula for the heat released by cooling:- **Cooling of Water:** \( q_\text{cool water} = m \cdot c \cdot \Delta T \), where \( \Delta T \) is the temperature change from 100掳C to 34掳C.Calculation:- \( q_\text{cool water} = 25.0 \text{ g} \times 4.186 \text{ J/(g路掳C)} \times 66 \text{ 掳C} = 6907.5 \text{ J} \)
04

Compare the Heat from Steam and Water

By comparing the heat quantities calculated: - Heat from steam (including condensation): 63407.5 J - Heat from water (just cooling): 6907.5 J The steam transfers much more heat to the skin, making steam burns potentially more severe than water burns of the same initial temperature and mass.

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Key Concepts

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

Latent Heat of Vaporization
When something changes from a liquid to a gas, it absorbs a large amount of energy. This energy is known as the "latent heat of vaporization." This concept is crucial in understanding why steam burns can be so harmful.

Latent heat of vaporization is the heat required to change one gram of a liquid into a gas at a constant temperature. For water, this value is particularly high at approximately 2260 J/g.
  • When steam, which is water vapor, comes into contact with skin, it releases this stored energy.
  • Even when steam just cools down, it releases a huge amount of energy during condensation, leading to severe burns.
Without factoring in latent heat, you cannot accurately estimate the heat released during steam burns.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is needed to raise the temperature of one gram of a substance by one degree Celsius. Understanding specific heat capacity can help make sense of why certain substances feel hot or cold when they transfer heat.

Water has a specific heat capacity of 4.186 J/(g路掳C), which is quite high. This means water changes temperature much more slowly compared to substances with lower specific heat values.
  • When water cools down from a high temperature, it releases heat gradually due to its high specific heat capacity.
  • This gradual release of heat can still contribute significantly to burns, but is minor when compared to the latent heat released upon steam condensation.
Specific heat capacity helps understand the slow temperature change and heating ability of water.
Phase Change
A phase change is when a substance changes from one state of matter to another, such as solid to liquid, or liquid to gas. During a phase change, you might notice that the temperature of the substance doesn鈥檛 change, even though heat is being added or removed.

When steam condenses into water, it's undergoing a phase change from gas to liquid, releasing latent heat in the process.
  • This is why steam burns are more severe than hot water burns; during a phase change, a lot of energy is transferred as heat to your skin.
  • Heat from phase changes can cause rapid temperature changes, leading to more intense burns.
Phase changes are pivotal in understanding heat transfer and the severity of burns caused by steaming substances.
Thermal Energy
Thermal energy is the total kinetic energy of the particles in a substance due to their motion. The faster particles move, the higher the temperature and thermal energy.

When steam or water transfers its thermal energy to your skin, it causes burns. More thermal energy transfer results in more severe burns.
  • Thermal energy from steam not only comes from its temperature but also from the latent heat released during condensation.
  • In water, only the kinetic energy from its temperature contributes to the burn.
In essence, understanding how thermal energy plays a role in heat transfer allows one to predict the effects of such burns, distinguishing the serious nature of steam burns from those of hot water.

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Most popular questions from this chapter

An ice-cule tray of negligible mass contains 0.350 \(\mathrm{kg}\) of water at \(18.0^{\circ} \mathrm{C} .\) How much heat must be removed to cool the water to \(0.00^{\circ} \mathrm{C}\) and freeze it? Express your answer in joules, calories, and Btu.

An electric kitchen range has a total wall area of 1.40 \(\mathrm{m}^{2}\) and is insulated with a layer of fiberglass 4.00 \(\mathrm{cm}\) thick. The inside surface of the fiberglass has a temperature of \(175^{\circ} \mathrm{C},\) and its outside surface is at \(35.0^{\circ} \mathrm{C}\) . The fiberglass has a thermal conductivity of 0.040 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 \(\mathrm{m}^{2} ?\) (b) What electric-power input to the heating element is required to maintain this temperature?

\(\mathrm{A} 500.0-\mathrm{g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 \(\mathrm{kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\) (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) What if the heat absorbed by the Styrofoam actually is not negligible. How would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

While painting the top of an antenna 225 \(\mathrm{m}\) in height, a worker accidentally lets a \(1.00-\mathrm{L}\) wattle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

A laboratory technician drops a \(0.0850-\mathrm{kg}\) sample of unknown solid material, at a temperature of \(100.0^{\circ} \mathrm{C},\) into a calorimeter. The calorimeter can, initially at \(19.0^{\circ} \mathrm{C},\) is made of 0.150 \(\mathrm{kg}\) of copper and contains 0.200 \(\mathrm{kg}\) of water. The final temperature of the calorimeter can and contents is \(26.1^{\circ} \mathrm{C}\) . Compute the specific heat of the sample.
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