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A heart defibrillator passes 10.0 A through a patient's torso for \(5.00 \mathrm{ms}\) in an attempt to restore normal beating. (a) How much charge passed? (b) What voltage was applied if \(500 \mathrm{J}\) of energy was dissipated? (c) What was the path's resistance? (d) Find the temperature increase caused in the \(8.00 \mathrm{kg}\) of affected tissue.

Short Answer

Expert verified
The charge passed through the patient's torso is 0.0500 C. The voltage applied during the process is 10,000 V. The path's resistance is 1,000 Ω. The temperature increase caused in the 8.00 kg of affected tissue is 1.50 x 10^(-2) K.

Step by step solution

01

(a) Calculate the charge passed through the patient's torso

First, we need to find the total charge that passed through the patient's torso. We use the formula \(Q = I \cdot t\), where \(I = 10.0 A\) and \(t = 5.00 ms\). \(Q = 10.0 A \cdot 5.00 \times 10^{-3} s = 0.0500 C\) So, the charge that passed through the patient's torso is 0.0500 C.
02

(b) Find the voltage applied during the process

Next, we need to find the voltage applied. We will use the formula \(V = \frac{W}{Q}\), where \(W = 500 J\) and \(Q = 0.0500 C\). \(V = \frac{500 J}{0.0500 C} = 10,000 V\) So, the voltage applied during the process is 10,000 V.
03

(c) Determine the path's resistance

Now, we will find the path's resistance using the formula \(R = \frac{V}{I}\), where \(V = 10,000 V\) and \(I = 10.0 A\). \(R = \frac{10,000 V}{10.0 A} = 1,000 \Omega\) So, the path's resistance is 1,000 Ω.
04

(d) Calculate the temperature increase in the affected tissue

Finally, we need to find the temperature increase in the affected tissue. We will use the formula \(\Delta T = \frac{W}{m \cdot c}\), where \(W = 500 J\), \(m = 8.00 kg\), and the specific heat capacity of the tissue is assumed to be that of water, with \(c = 4190 \frac{J}{kg \cdot K}\). \(\Delta T = \frac{500 J}{8.00 kg \cdot 4190 \frac{J}{kg \cdot K}} = 1.50 \times 10^{-2} K\) So, the temperature increase caused in the 8.00 kg of affected tissue is 1.50 x 10^(-2) K.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that explains how objects can attract or repel each other. To imagine how a defibrillator works, think of a heart like a malfunctioning engine that needs a jump-start. The defibrillator delivers a quick electric shock to the heart, which is essentially a charge moving from one place to another.

In our exercise, the charge is calculated using the formula:
\(Q = I \cdot t\)
Where \(I\) is the current in amperes (A) and \(t\) is the time in seconds (s). In the exercise, we found that a current of 10.0 A passing for 5.00 milliseconds (which is 5.00 x 10^-3 seconds) results in a charge of 0.0500 coulombs (C). This charge is the 'push' needed to potentially restart the heart's normal rhythm.
Electric Voltage
The concept of electric voltage is often compared to the pressure in a water hose; it's the force that pushes the electric charge through a circuit. Voltage is what a defibrillator relies on to actually get the charge through the body's natural resistance to a patient's heart.

Using the formula \(V = \frac{W}{Q}\), we can understand voltage (\(V\)) as the amount of energy (\(W\text{, in joules}\)) per unit charge (\(Q\text{, in coulombs}\)). As the solution to the exercise demonstrates, if 500 joules of energy is used to move 0.0500 coulombs of charge, the voltage applied is a substantial 10,000 volts. It's this strong voltage that allows the charge to penetrate through the tissue and reach the heart.
Electrical Resistance
Just as the body has physical muscles that resist external forces, electrical resistance (\(R\text{, in ohms}\)) is the opposition to the flow of electric charge. It's like running against a strong wind—it takes more effort (or voltage) to push forward. A defibrillator must overcome this resistance to deliver its life-saving charge.

By using the formula \(R = \frac{V}{I}\), we decipher the resistance encountered as we divide the voltage by the current. The exercise tells us the defibrillator faced 1,000 ohms of resistance, a measure of how much the body restricts the charge's journey.
Specific Heat Capacity
Warm up a pot of water and you're experiencing specific heat capacity, the amount of heat required to raise the temperature of a substance. Here, it's not about cooking but understanding how the defibrillator's energy affects body tissue.

The formula \(\Delta T = \frac{W}{m \cdot c}\) relates the energy (\(W\)) to the mass of the tissue (\(m\)) and its specific heat capacity (\(c\)). Using water's specific heat capacity shows how even a small amount of energy can heat body tissue, in this case causing a temperature increase of 0.015 degrees Celsius, as per our exercise solution. These calculations are crucial for ensuring that the defibrillator safely restores normal heart rhythm without damaging tissues.

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Most popular questions from this chapter

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