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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

You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is \(28.4 \mathrm{~N}\). You carefully add \(1.25 \times 10^{4} \mathrm{~J}\) of heat energy to the sample and find that its temperature rises \(18.0 \mathrm{C}^{\circ} .\) What is the sample's specific heat?

One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\). To treat a \(70.0 \mathrm{~kg}\) patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C}\) ) you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat of the human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{C}^{\circ},\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C}\).

Consider a poor lost soul walking at \(5 \mathrm{~km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of \(280 \mathrm{~W},\) and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\mathrm{skin}}\left(T_{\mathrm{air}}-T_{\mathrm{skin}}\right),\) where \(k^{\prime}\) is \(54 \mathrm{~J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2},\) the exposed skin area \(A_{\text {skin }}\) is \(1.5 \mathrm{~m}^{2},\) the air temperature \(T_{\text {air }}\) is \(47^{\circ} \mathrm{C},\) and the skin temperature \(T_{\text {skin }}\) is \(36^{\circ} \mathrm{C} ;\) (iii) the skin absorbs radiant energy from the sun at a rate of \(1400 \mathrm{~W} / \mathrm{m}^{2} ;\) (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C}\). (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C}\). Which mechanism is the most important? (b) At what rate (inL/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) ) (c) Suppose the person is protected by light-colored clothing \((e \approx 0)\) and only \(0.45 \mathrm{~m}^{2}\) of skin is exposed. What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

The emissivity of tungsten is 0.350 . A tungsten sphere with radius \(1.50 \mathrm{~cm}\) is suspended within a large evacuated enclosure whose walls are at \(290.0 \mathrm{~K}\). What power input is required to maintain the sphere at \(3000.0 \mathrm{~K}\) if heat conduction along the supports is ignored?

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

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