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Assuming the severity of a burn increases as the amount of energy put into the skin increases, which of the following would cause the most severe burn (assume equal masses)? a) water at \(90^{\circ} \mathrm{C}\) b) copper at \(110^{\circ} \mathrm{C}\) c) steam at \(180^{\circ} \mathrm{C}\) d) aluminum at \(100^{\circ} \mathrm{C}\) e) lead at \(100^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: (c) steam at \(180^{\circ} \mathrm{C}\)

Step by step solution

01

Research the specific heat capacities of the materials

We need to look up the specific heat capacities of each material in order to make a comparison. Here are the specific heat capacities of each material listed in the problem: Water: \(4.18 \ \mathrm{J / (g \cdot K)}\) Copper: \(0.385 \ \mathrm{J / (g \cdot K)}\) Steam: \(2.03 \ \mathrm{J / (g \cdot K)}\) Aluminum: \(0.897 \ \mathrm{J / (g \cdot K)}\) Lead: \(0.127 \ \mathrm{J / (g \cdot K)}\)
02

Decide on the temperature difference of each material to skin

Next, we need to decide on the temperature difference for each material. We will assume that the temperature of the skin is \(30^{\circ} \mathrm{C}\). The temperature difference (\(ΔT\)) for each material is: Water: \(60^{\circ} \mathrm{C}\) Copper: \(80^{\circ} \mathrm{C}\) Steam: \(150^{\circ} \mathrm{C}\) Aluminum: \(70^{\circ} \mathrm{C}\) Lead: \(70^{\circ} \mathrm{C}\)
03

Calculate the total energy absorbed by skin for each material

Now, we will calculate the energy absorbed by the skin for each material using the specific heat capacities and temperature differences found in steps 1 and 2: Water: \(Q = 4.18 \times 60 = 250.8 \ \mathrm{J}\) Copper: \(Q = 0.385 \times 80 = 30.8 \ \mathrm{J}\) Steam: \(Q = 2.03 \times 150 = 304.5 \ \mathrm{J}\) Aluminum: \(Q = 0.897 \times 70 = 62.79 \ \mathrm{J}\) Lead: \(Q = 0.127 \times 70 = 8.89 \ \mathrm{J}\)
04

Compare the energy values

We can now compare the calculated energy values to determine which material will cause the most severe burn: Water: \(250.8 \ \mathrm{J}\) Copper: \(30.8 \ \mathrm{J}\) Steam: \(304.5 \ \mathrm{J}\) Aluminum: \(62.79 \ \mathrm{J}\) Lead: \(8.89 \ \mathrm{J}\) We can conclude that, given equal masses, steam at \(180^{\circ} \mathrm{C}\) would cause the most severe burn, as the energy absorbed by the skin is the highest (\(304.5 \ \mathrm{J}\)) among all materials. Therefore, the correct answer is (c) steam at \(180^{\circ} \mathrm{C}\).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a property that describes how much heat energy is needed to raise the temperature of a certain mass of substance by one degree Celsius (or one Kelvin). Formally, it is expressed as joules per gram per degree Celsius \( \mathrm{J / (g \cdot ^{\circ}C)} \).

A substance with a high specific heat capacity can absorb a lot of heat without changing its temperature significantly. This is crucial when considering burn severity, as a material's specific heat capacity influences the amount of energy it can transfer to the skin. For instance, water has a high specific heat capacity \(4.18 \mathrm{J / (g \cdot K)} \), which means it can store and release a large amount of energy, potentially causing more severe burns compared to other materials with a lower specific heat capacity, such as lead \(0.127 \mathrm{J / (g \cdot K)} \).

Understanding specific heat capacity helps us to predict the outcome when different materials at various temperatures come into contact with the skin, by showing how much energy can be transferred and possibly causing burns.
Thermal Energy Transfer
Thermal energy transfer, often referred to as heat transfer, describes the flow of heat energy from a hotter object to a cooler one. There are three methods of heat transfer: conduction, convection, and radiation. In the context of burn severity, we primarily consider conduction, which is the direct transfer of heat through a medium or between objects in physical contact.

If a material at a high temperature comes into contact with skin, thermal energy will be transferred until there is no temperature difference between the two, hence the severity of a burn highly depends on the amount of energy transferred. In the exercise, materials with higher temperatures and specific heat capacities have the potential to transfer more energy to the skin. For example, steam at \(180^\circ \mathrm{C}\) can transfer substantial energy leading to severe burns, particularly because steam can also release latent heat during condensation.
Temperature Difference
Temperature difference, often symbolized as \( \Delta T \), is a critical factor in the severity of burns due to its direct impact on the amount of thermal energy transferred. It's the difference in temperature between two substances - in this case, between the skin and the material in question. The larger the temperature difference, the higher the rate of thermal energy transfer, at least initially.

In our exercise, assuming the skin has a temperature of \(30^\circ \mathrm{C}\), materials with higher initial temperatures like steam will have a larger \( \Delta T \). Hence, they can initially transfer energy more rapidly compared to materials with temperatures closer to that of the skin. The equation for heat transfer used in the step by step solution \(Q = m \cdot c \cdot \Delta T\) takes into account this temperature difference, which is why steam, despite its lower specific heat capacity than water, can cause more severe burns due to its much higher temperature relative to the skin.

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Most popular questions from this chapter

A gas enclosed in a cylinder by means of a piston that can move without friction is warmed, and 1000 J of heat enters the gas. Assuming that the volume of the gas is constant, the change in the internal energy of the gas is a) 0 . b) 1000 J. c) -1000 J. d) none of the above.

Why is a dry, fluffy coat a better insulator than the same coat when it is wet?

An aluminum block of mass \(M_{\mathrm{Al}}=2.0 \mathrm{~kg}\) and specific heat \(C_{\mathrm{Al}}=910 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\) is at an initial temperature of \(1000{ }^{\circ} \mathrm{C}\) and is dropped into a bucket of water. The water has mass \(M_{\mathrm{H}_{2} \mathrm{O}}=12 \mathrm{~kg}\) and specific heat \(C_{\mathrm{H}_{2} \mathrm{O}}=4190 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\) and is at room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) What is the approximate final temperature of the system when it reaches thermal equilibrium? (Neglect heat loss out of the system.) a) \(50^{\circ} \mathrm{C}\) b) \(60^{\circ} \mathrm{C}\) c) \(70^{\circ} \mathrm{C}\) d) \(80^{\circ} \mathrm{C}\)

You were lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of 1.00 . b) Calculate the net radiated power from your body when you were inside a shelter at \(20.0^{\circ} \mathrm{C}\). c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\).

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

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