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

Short Answer

Expert verified
The total heat transfer from the steam is \(63403 \mathrm{~J}\) and from water is \(6896 \mathrm{~J}\). This indicates that steam causes more serious burns than water at the same temperature because of the additional heat released during the process of condensation.

Step by step solution

01

Heat Release by Condensation

Calculate the amount of heat released by steam during condensation using the heat of fusion formula: \(Q = mL_f\), where \(m\) is the mass of the substance (steam) in grams, and \(L_f\) is the latent heat of fusion of water, which is \(2260 \mathrm{~J/g}\) at \(100.0^{\circ} \mathrm{C}\). Therefore, the heat released by the condensation of \(25.0 \mathrm{~g}\) of steam at \(100.0^{\circ} \mathrm{C}\) can be found by substituting the known values into the formula: \(Q = 25.0 \mathrm{~g} \times 2260 \mathrm{~J/g} = 56500 \mathrm{~J}\)
02

Heat Loss from Steam

Next, calculate the amount of heat lost from the steam as it cools to skin temperature using the formula \(Q = mc\Delta T\). In place of \(c\) insert the specific heat of water, \(4.186 \mathrm{~J/g\cdot C}\), \(m = 25.0 \mathrm{~g}\), and \(\Delta T = 100.0^{\circ} \mathrm{C} - 34.0^{\circ} \mathrm{C}\). Therefore, plugging in these values will yield the total heat loss from the steam: \(Q = 25.0 \mathrm{~g} \times 4.186 \mathrm{~J/g\cdot C} \times 66.0^{\circ} \mathrm{C} = 6903 \mathrm{~J}\). The total heat absorbed by the skin is the sum of the heat released during condensation and the heat lost due to cooling: \(56500 \mathrm{~J} + 6903 \mathrm{~J} = 63403 \mathrm{~J}\).
03

Heat Loss from Water

Finally, calculate the amount of heat transferred from the \(25.0 \mathrm{~g}\) of water which cools from \(100.0^{\circ} \mathrm{C}\) to skin temperature, \(34.0^{\circ} \mathrm{C}\), using the same formula for heat transfer (\(Q = mc\Delta T\)) as previously mentioned. Using the specific heat of water, the mass of water, and the change in temperature, the total heat loss from the water is: \(Q = 25.0 \mathrm{~g} \times 4.186 \mathrm{~J/g\cdot C} \times (100.0^{\circ} \mathrm{C} - 34.0^{\circ} \mathrm{C}) = 6896 \mathrm{~J}\).
04

Comparison

By comparing the total heat absorbed by the skin from the steam (\(63403 \mathrm{~J}\)) and the heat absorbed from the water (\(6896 \mathrm{~J}\)), it can be concluded that a significantly greater amount of heat is absorbed from steam, which implies that burns from steam are more severe than burns from water at the same temperature

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Key Concepts

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

Latent Heat of Fusion
When a substance changes from one state of matter to another—like from a solid to a liquid—the energy involved in this phase transition is called the latent heat of fusion. This is often simply termed as the heat of fusion and is unique for each substance. For instance, to convert water to ice or ice to water, energy must either be absorbed or released by water molecules.

In our exercise problem, the heat of fusion comes into play when steam at 100 degrees Celsius condenses into water without changing the temperature. It's this hidden energy, which does not affect the temperature, that contributes significantly to the thermal energy transferred. The formula to calculate this energy is given by \(Q = mL_f\), where \(m\) is the mass of the substance undergoing the phase change, and \(L_f\) is the latent heat of fusion. For water at its boiling point, \(L_f\) is 2260 J/g; thus steam releasing latent heat can cause severe burns, more so than water at the same temperature.
Specific Heat Capacity
The specific heat capacity, denoted as \(c\), is a property that describes how much heat energy (in joules) is needed to raise the temperature of one gram of a substance by one degree Celsius. This concept helps us understand how different materials react to heat. For example, metals typically have a low specific heat capacity, which means they heat up and cool down quickly.

In the context of our exercise, the specific heat capacity of water is used in the formula \(Q = mc\Delta T\) to calculate the heat lost as steam and hot water cool down to skin temperature. This formula quantifies the thermal energy (\(Q\)) exchanged during a temperature change (\(\Delta T\)) by accounting for the material's mass (\(m\)) and its ability to store heat (\(c\)). Water has a relatively high specific heat capacity of 4.186 J/g°C, which means it can hold a significant amount of heat. This characteristic of water is essential for understanding the heat exchange process with the skin in the given scenarios.
Thermal Energy in Heat Exchange
Thermal energy in heat exchange refers to the transfer of heat between objects or systems resulting from a temperature difference. In essence, heat flows from a higher temperature object to a lower temperature one until thermal equilibrium is reached. The amount of thermal energy transferred is influenced by the properties of the substances involved, such as their mass, specific heat capacity, and phase changes.

In our exercise, thermal energy transfer occurs when steam and hot water are cooled to the skin's temperature. The presence of steam entails two parts of heat transfer: the release of latent heat during the phase change from steam to water, and the subsequent cooling of the resultant water. Conversely, the hot water's cooling involves only the specific heat exchange with no latent heat, as it's already in the liquid phase. We can quantify the thermal energy transferred with the formula \(Q = mc\Delta T\) for the cooling process and \(Q = mL_f\) for the phase change, elucidating the more substantial impact steam has due to the additional latent heat release. This can explain why steam burns can be much more severe compared to hot water burns.

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Most popular questions from this chapter

A carpenter builds an exterior house wall with a layer of wood \(3.0 \mathrm{~cm}\) thick on the outside and a layer of Styrofoam insulation \(2.2 \mathrm{~cm}\) thick on the inside wall surface. The wood has \(k=0.080 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K},\) and the Styrofoam has \(k=0.027 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) The interior surface temperature is \(19.0^{\circ} \mathrm{C},\) and the exterior surface temperature is \(-10.0^{\circ} \mathrm{C}\). (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

Animals in cold climates often depend on \(t w o\) layers of insulation: a layer of body fat (of thermal conductivity \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere \(1.5 \mathrm{~m}\) in diameter having a layer of fat \(4.0 \mathrm{~cm}\) thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at \(2.7^{\circ} \mathrm{C}\) during hibernation, while the inner surface of the fat layer is at \(31.0^{\circ} \mathrm{C}\). (a) What is the temperature at the fat-inner fur boundary so that the bear loses heat at a rate of \(50.0 \mathrm{~W} ?\) (b) How thick should the air layer (contained within the fur) be?

There is \(0.050 \mathrm{~kg}\) of an unknown liquid in a plastic container of negligible mass. The liquid has a temperature of \(90.0^{\circ} \mathrm{C}\). To measure the specific heat capacity of the unknown liquid, you add a mass \(m_{w}\) of water that has a temperature of \(0.0^{\circ} \mathrm{C}\) to the liquid and measure the final temperature \(T\) after the system has reached thermal equilibrium. You repeat this measurement for several values of \(m_{\mathrm{w}},\) with the initial temperature of the unknown liquid always equal to \(90.0^{\circ} \mathrm{C}\). The plastic container is insulated, so no heat is exchanged with the surroundings. You plot your data as \(m_{\mathrm{w}}\) versus \(T^{-1}\), the inverse of the final temperature \(T\). Your data points lie close to a straight line that has slope \(2.15 \mathrm{~kg} \cdot \mathrm{C}^{\circ} .\) What does this result give for the value of the specific heat capacity of the unknown liquid?

Size of a Light-Bulb Filament. The operating temperature of a tungsten filament in an incandescent light bulb is \(2450 \mathrm{~K},\) and its emissivity is \(0.350 .\) Find the surface area of the filament of a \(150 \mathrm{~W}\) bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation appears as visible light.)

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

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