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Chapter 14: Problem 3

(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 mm thick. Assuming the total surface area of the wetsuit covering the diver is about 1.0 m\(^2\), and that ocean water enters the suit at 10\(^\circ\)C and is warmed by the diver to skin temperature of 35\(^\circ\)C, estimate how much energy (in units of candy bars \(=\) 300 kcal) is required by this heating process.

Short Answer

Expert verified
The diver needs approximately 0.042 candy bars of energy to heat the water in the wetsuit.

Step by step solution

01

Understand the problem

First, identify the information given and what is being asked. We have a wetsuit with a water layer, initial and final temperatures, and need to find the energy needed to heat the water, given in candy bars.
02

Identify relevant formulas and concepts

We need the formula for heat energy, \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the temperature change. We will convert the energy into kcal for comparison with candy bar calories.
03

Calculate the volume of water

Multiply the surface area of the wetsuit by the thickness of the water layer to find the volume. \[ V = 1.0 \, \text{m}^2 \times 0.5 \, \text{mm} = 1.0 \, \text{m}^2 \times 0.0005 \, \text{m} = 0.0005 \, \text{m}^3 \]
04

Determine the mass of the water

Use the density of water (1000 kg/m\(^3\)) to find the mass. \[ m = \text{density} \times \text{volume} = 1000 \, \text{kg/m}^3 \times 0.0005 \, \text{m}^3 = 0.5 \, \text{kg} \]
05

Calculate the energy required for heating

We know the specific heat capacity of water is \(4.18 \, \text{J/g}^\circ \text{C}\), but it needs to be in kg. We then apply it in the formula. \[ Q = mc\Delta T = 0.5 \, \text{kg} \times 4180 \, \frac{\text{J}}{\text{kg}^\circ \text{C}} \times (35 - 10)^\circ \text{C} \]\[ Q = 0.5 \times 4180 \times 25 = 52250 \, \text{J} \]
06

Convert the energy to kcal

Convert joules to calories first (1 cal = 4.184 J) and then to kilocalories.\[ Q = \frac{52250}{4.184} = 12486.63 \, \text{cal} \]\[ Q = \frac{12486.63}{1000} = 12.49 \, \text{kcal} \]
07

Convert kilocalories to candy bars

Use the given conversion factor where 1 candy bar = 300 kcal.\[ \text{Candy bars} = \frac{12.49}{300} = 0.04163 \approx 0.042 \]

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Key Concepts

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

Energy Conversion
In the context of the diver heating water inside a wetsuit, energy conversion plays a vital role. When the diver's body generates heat, it effectively converts chemical energy stored in the body into thermal energy, which then warms the surrounding water. The process can be seen as an energy conversion chain:
  • Chemical Energy: The diver derives energy from food converted into usable energy via metabolic processes.
  • Thermal Energy: This chemical energy is then transformed into heat as the body works to maintain a stable temperature.
  • Heat Transfer: The thermal energy raises the temperature of the cooler water around the diver.
Energy conversion in this instance is crucial for maintaining a warm environment in the cold ocean, ensuring the diver's comfort and safety. Understanding this principle helps us appreciate how energy flows and transforms in practical scenarios.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or material to another. In this exercise, heat transfer occurs when warm body heat from the diver raises the temperature of the cooler water inside the wetsuit. There are three primary modes of heat transfer:
  • Conduction: Thermal energy moves through direct contact, e.g., from the diver's skin to the adjacent water layer. This is the primary mode at play here.
  • Convection: If the water could circulate, it would distribute the heat, but in the tightly sealed layer of the wetsuit, convection minimizes.
  • Radiation: Less significant here, heat radiation could occur, but the direct contact makes conduction dominant.
The calculation part of the solution involved assessing how much energy was needed to raise the temperature of the water. This helps students understand how energy values translate to real-world heating processes.
Wetsuit Insulation
Wetsuit insulation is vital for divers spending time in cold water. It functions by creating a barrier that helps retain body heat while minimizing heat loss to the water. The principle behind wetsuit insulation relies on trapping a thin layer of water between the suit and the skin.
  • Water Layer Insulation: This minimal water layer is initially cold but is quickly heated by body heat.
  • Neoprene Material: Wetsuits are made of neoprene, which provides further insulation by reducing thermal conductivity.
  • Thermal Equilibrium: Once the body heats the water, it forms a thermal equilibrium, maintaining a warmer microenvironment.
The effectiveness of wetsuit insulation largely depends on the right fit, thickness of the material, and water temperature. Understanding these concepts enables divers to optimize their protection against cold environments efficiently.

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Most popular questions from this chapter

(II) High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from your body if you: (\(a\)) eat 1.0 kg of \(-15^\circ\)C snow which your body warms to body temperature of 37\(^\circ\)C; (\(b\)) melt 1.0 kg of \(-15^\circ\)C snow using a stove and drink the resulting 1.0 kg of water at 2\(^\circ\)C, which your body has to warm to 37\(^\circ\)C.

(II) An average active person consumes about 2500 Cal a day. (\(a\)) What is this in joules? (\(b\)) What is this in kilowatthours? (\(c\)) If your power company charges about 10 cents per kilowatt-hour, how much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much money per day?

A soft-drink can contains about 0.35 kg of liquid at 5\(^\circ\)C. Drinking this liquid can actually consume some of the fat in the body, since energy is needed to warm the liquid to body temperature (37\(^\circ\)C). How many food Calories should the drink have so that it is in perfect balance with the heat needed to warm the liquid (essentially water)?

(II) The \(heat \, capacity\), \(C\), of an object is defined as the amount of heat needed to raise its temperature by 1 C\(^\circ\). Thus, to raise the temperature by requires heat \(Q\) given by \(\Delta T\) $$Q = C \, \Delta \, T$$ (\(a\)) Write the heat capacity C in terms of the specific heat, \(c\), of the material. (\(b\)) What is the heat capacity of 1.0 kg of water? (\(c\)) Of 45 kg of water?

(II) Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of \(-39.0^\circ\)C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80\(^\circ\)C; the resulting equilibrium temperature is 5.06\(^\circ\)C.
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