/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 (II) When a diver jumps into the... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 2

(II) When a diver jumps into the ocean, water leaks into the gap region between the diver's skin and her wetsuit, forming a water layer about \(0.5 \mathrm{~mm}\) thick. Assuming the total surface area of the wetsuit covering the diver is about \(1.0 \mathrm{~m}^{2},\) and that ocean water enters the suit at \(10^{\circ} \mathrm{C}\) and is warmed by the diver to skin temperature of \(35^{\circ} \mathrm{C},\) estimate how much energy (in units of candy bars \(=300 \mathrm{kcal}\) ) is required by this heating process.

Short Answer

Expert verified
The energy required is approximately 0.042 candy bars.

Step by step solution

01

Determine Water Mass

First, calculate the mass of the water layer. The volume of water can be found using the surface area of the wetsuit and the thickness of the water layer. The volume \( V \) is determined by \( V = ext{Area} \times ext{Thickness} = 1.0 ext{ m}^2 \times 0.5 \times 10^{-3} ext{ m} \). This gives us \( V = 0.0005 ext{ m}^3 \). Given that the density of water is approximately \( 1000 ext{ kg/m}^3 \), the mass \( m \) of the water is \( m = ext{Density} \times ext{Volume} = 1000 ext{ kg/m}^3 \times 0.0005 ext{ m}^3 = 0.5 ext{ kg} \).
02

Calculate Energy Required to Heat Water

Next, calculate the energy required to raise the temperature of the water from \(10^{\circ}C\) to \(35^{\circ}C\). The formula for heat energy \( Q \) required is \( Q = m \cdot c \cdot \Delta T \), where \( m \) is the mass of water, \( c \) is the specific heat capacity of water \( (4.186 ext{ J/g掳C or } 4186 ext{ J/kg掳C}) \), and \( \Delta T \) is the change in temperature \( (35 - 10 = 25掳C) \). Thus, \( Q = 0.5 \text{ kg} \times 4186 \text{ J/kg掳C} \times 25掳C = 52325 \text{ J} \).
03

Convert Energy to Calories

Convert the energy from joules to calories, knowing that \(1 ext{ cal} = 4.186 ext{ J}\). Therefore, the energy in calories \( Q_{cal} = \frac{52325 ext{ J}}{4.186 ext{ J/cal}} = 12500 ext{ cal} \). Since \(1 ext{ kcal} = 1000 ext{ cal}\), we have \( Q_{kcal} = 12.5 ext{ kcal} \).
04

Calculate Energy in Candy Bars

Finally, calculate how many candy bars worth of energy are required. Since one candy bar is equivalent to \(300 ext{ kcal}\), the number of candy bars is given by \( \frac{12.5 ext{ kcal}}{300 ext{ kcal/candy bar}} = 0.0417 \text{ candy bars} \). Round this to a reasonable amount, approximately \(0.042 \text{ candy bars}\).

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Key Concepts

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

Heat Transfer
Heat transfer is an essential concept in thermodynamics, referring to the movement of thermal energy between objects or substances due to a temperature difference. In the ocean diving scenario, when the diver enters the water, there is a temperature difference between the diver's body (skin temperature) and the incoming ocean water.
This temperature difference drives the flow of heat, causing the diver's body to warm up the ocean water that leaks into the wetsuit.
  • Heat always moves from a warmer object (the diver) to a cooler one (the ocean water) until equilibrium is reached.
  • The heat transfer continues until the water in the wetsuit reaches the temperature of the diver's skin.
Understanding how heat transfers in different scenarios helps us solve problems related to energy requirements, as seen in this context where we estimate the energy needed to warm up the water layer.
Specific Heat Capacity
Specific heat capacity is a property of a material that defines how much heat energy is required to change the temperature of a unit mass of a substance by one degree Celsius. It's a critical factor in calculating thermal energy change.
Water has a relatively high specific heat capacity, which is why it takes a considerable amount of energy to change its temperature.
  • The specific heat capacity of water is approximately 4186 J/kg掳C.
  • This high specific heat means that even a small volume of water can absorb or release a significant amount of heat with a minor change in temperature.
Within our ocean diving example, this property of water significantly impacts the amount of thermal energy the diver's body has to supply to raise the water's temperature by 25掳C. This emphasizes how the specific heat capacity impacts practical issues such as thermal comfort in different environments.
Thermal Energy Calculation
Calculating thermal energy involves understanding how changes in temperature require energy. In the case of the diver and the ocean water, we use the formula for calculating heat energy:\[ Q = m \cdot c \cdot \Delta T \]where:
  • \(Q\) is the heat energy (in joules or calories),
  • \(m\) is the mass of the water,
  • \(c\) is the specific heat capacity,
  • \(\Delta T\) is the temperature change (in 掳C).
For our problem:
  • The mass \(m\) of the water is 0.5 kg.
  • The specific heat capacity \(c\) is 4186 J/kg掳C.
  • The temperature change \(\Delta T\) is 25掳C.
By substituting these values into the formula, we determine the energy required to heat the water. Moreover, by converting energy from joules to other units like calories or candy bars, we can better grasp the practical energy equivalence, making the concept more relatable in terms of everyday energy consumption.

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Most popular questions from this chapter

A scuba diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of \(14.0 \mathrm{~m}\). Assume the temperature is constant at \(298 \mathrm{~K}\), and that the air behaves as an ideal gas. (a) How large is the bubble when it reaches the surface? (b) Sketch a \(P V\) diagram for the process. (c) Apply the first law of thermodynamics to the bubble, and find the work done by the air in rising to the surface, the change in its internal energy, and the heat added or removed from the air in the bubble as it rises. Take the density of water to be \(1000 \mathrm{~kg} / \mathrm{m}^{3}\)

(II) A \(31.5-\mathrm{g}\) glass thermometer reads \(23.6^{\circ} \mathrm{C}\) before it is placed in 135 \(\mathrm{mL}\) of water. When the water and thermometer come to equilibrium, the thermometer reads \(39.2^{\circ} \mathrm{C}\) . What was the original temperature of the water? [Hint: Ignore the mass of fluid inside the glass thermometer.]

(II) When a \(290-\) g piece of iron at \(180^{\circ} \mathrm{C}\) is placed in a 95 -g aluminum calorimeter cup containing 250 \(\mathrm{g}\) of glycerin at \(10^{\circ} \mathrm{C},\) the final temperature is observed to be \(38^{\circ} \mathrm{C}\) Estimate the specific heat of glycerin.

(II) A \(58-\mathrm{kg}\) ice-skater moving at \(7.5 \mathrm{~m} / \mathrm{s}\) glides to a stop. Assuming the ice is at \(0^{\circ} \mathrm{C}\) and that \(50 \%\) of the heat generated by friction is absorbed by the ice, how much ice melts?

(II) An audience of 1800 fills a concert hall of volume \(22,000 \mathrm{~m}^{3}\). If there were no ventilation, by how much would the temperature of the air rise over a period of \(2.0 \mathrm{~h}\) due to the metabolism of the people ( 70 W/person)?
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