/*! 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 79 Struck by lightning. Lightning s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Struck by lightning. Lightning strikes can involve currents as high as \(25,000\) A that last for about 40\(\mu\) s. If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is 75 kg, that he is wet (after all, he is in a rainstorm) and therefore has a resistance of \(1.0 \mathrm{k} \Omega,\) and that his body is all water (which is reasonable for a rough, but plausible, approximation).(a) By how many degrees Celsius would this lightning bolt increase the temperature of 75 kg of water? (b) Given that the internal body temperature is about \(37^{\circ} \mathrm{C}\) , would the person's temperature actually increase that much? Why not? What would happen first?

Short Answer

Expert verified
The temperature would increase by approximately 79.63°C, but severe damage would occur before this thermal increase.

Step by step solution

01

Calculate Energy Delivered

To find the energy delivered to the person's body, we use the formula for electrical energy: \[ E = I^2 \cdot R \cdot t \]where \( I = 25,000 \) A is the current, \( R = 1,000 \) \( \Omega \) (since 1 k\( \Omega = 1,000 \Omega \)) is the resistance, and \( t = 40 \mu s = 40 \times 10^{-6} \) s is the time duration. Let's calculate:\[ E = (25,000)^2 \cdot 1,000 \cdot 40 \times 10^{-6} \]\[ E = 25,000^2 = 625,000,000 \]\[ E = 625,000,000 \cdot 1,000 \cdot 40 \times 10^{-6} = 25,000,000 \text{ J} \]Thus, the energy delivered is 25,000,000 J.
02

Calculate Temperature Increase

Now, we use the formula for heat transferred to a substance:\[ Q = mc\Delta T \]where \( Q = 25,000,000 \text{ J} \) from Step 1, \( m = 75 \text{ kg} \) is the mass of the water (person's body), and \( c = 4,186 \text{ J/kg°C} \) is the specific heat capacity of water. We solve for \( \Delta T \):\[ \Delta T = \frac{Q}{mc} = \frac{25,000,000}{75 \times 4,186} \]\[ \Delta T = \frac{25,000,000}{313950} \approx 79.63°C \]Thus, the temperature increase would be approximately 79.63°C.
03

Analyze Physical Effects

Given the calculated increase of approximately 79.63°C, the internal body temperature would rise far beyond survivable levels (from 37°C to over 116°C). However, in reality, this increase would not occur because such a sudden and high magnitude current would cause immediate severe burns and other injuries, likely leading to immediate cardiac arrest or severe tissue damage before such a temperature rise could be achieved. Hence, the person would likely suffer fatal injuries from the electrical discharge rather than the thermal increase.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Lightning Strike Analysis
When analyzing a lightning strike, we deal with colossal amounts of electrical energy being transferred in a very short timeframe. Lightning strikes can generate currents as high as 25,000 A, lasting merely for fractions of a second, around 40 microseconds ( ext{µs} ). To put this in perspective, household electrical currents are typically around 15 to 20 A. A lightning strike enters through one part of the body and exits through another, effectively using the human body as a conductor. The body's resistance, with an assumed wet external surface, is approximately 1 ext{k} Ω (1,000 Ω). Resistance is crucial because it determines how much of this mammoth current can flow through the body, impacting the energy that the body absorbs. Understanding these parameters helps us to calculate the potential energy absorbed by a person during a lightning strike. This involves breaking down the concept of electrical energy transfer into calculations that yield the magnitude of energy involved.
Specific Heat Capacity
Specific heat capacity is a fundamental concept in understanding how substances absorb and retain thermal energy. It represents the amount of energy required to raise the temperature of a unit mass of a substance by 1°C. For instance: - Water has a high specific heat capacity, about 4,186 ext{J/kg°C} , meaning it can absorb a lot of heat before its temperature increases significantly. When calculating the effects of a lightning strike on the body, we assume the body is essentially a mass of water. This simplification lets us use the specific heat capacity of water to determine how much the body’s temperature would rise given certain levels of absorbed energy. A higher specific heat capacity is protective in controlling drastic temperature changes, thereby delaying harmful effects due to thermal energy absorption. This property is crucial in the case of a lightning strike, as it helps predict potential internal temperature changes.
Thermal Energy Calculation
Thermal energy calculations provide insights into the temperature increase resulting from energy absorbed from a lightning strike. We utilize the thermal energy formula: \[ Q = mc\Delta T \]where \( Q \) is the thermal energy added, \( m \) is the mass (in this case, 75 kg for the human body assumed to be water), \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.Plugging in the numbers obtained from the lightning calculation:- \( Q = 25,000,000 \) J (from the energy delivered in the lightning strike),- \( m = 75 \) kg,- \( c = 4,186 \text{ J/kg°C} \).The result is an estimated temperature rise of approximately 79.63°C. It is crucial to understand the real-world implications of this result. Despite the theoretical temperature rise, in practice, the body would experience immediate damage from the electrical current long before such a temperature shift occurred. This is due to the rapid and intense nature of the energy input which could cause burns and other severe injuries immediately upon exposure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A piece of wire has a resistance \(R\) . It is cut into three pieces of equal length, and the pieces are twisted together parallel to each other. What is the resistance of the resulting wire in terms of \(R ?\)

Calculate the (a) maximum and (b) minimum values of resistance that can be obtained by combining resistors of \(36 \Omega,\) \(47 \Omega,\) and 51\(\Omega .\)

A \(540-\mathrm{W}\) electric heater is designed to operate from 120 \(\mathrm{V}\) lines. (a) What is its resistance? (b) What current does it draw? (c) If the line voltage drops to \(110 \mathrm{V},\) what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.) (d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

A 4600\(\Omega\) resistor is connected across a charged 0.800 \(\mathrm{nF}\) capacitor. The initial current through the resistor, just after the connection is made, is measured to be 0.250 A. (a) What magnitude of charge was initially on each plate of this capacitor? (b) How long after the connection is made will it take before the charge is reduced to 1\(/ e\) of its maximum value?

Current in the body. The resistance of the body varies from approximately 500 \(\mathrm{k} \Omega\) (when it is very dry) to about 1 \(\mathrm{k} \Omega\) (when it is wet). The maximum safe current is about 5.0 \(\mathrm{mA} .\) At 10 \(\mathrm{mA}\) or above, muscle contractions can occur that may be fatal. What is the largest potential difference that a person can safely touch if his body is wet? Is this result within the range of common household voltages?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.