Problem 49 - College Physics

Question

A A nuclear device containing 8.00 kg of plutonium explodes. The rest mass of the products of the explosion is less than the original rest mass by one part in \(10^{4} .(\) a) How much energy is released in the explosion? (b) If the explosion takes place in \(4.00 \mu s,\) what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of 1.00 \(\mathrm{km}\) ?

Step-by-Step Solution

Answer

(a) Energy released is \(7.2 \times 10^{13} \text{ J}\). (b) Average power is \(1.8 \times 10^{19} \text{ W}\). (c) Mass of water lifted is \(7.34 \times 10^9 \text{ kg}\).

Step 1: Calculate the Mass Defect

First, find the mass defect due to the reaction. Since the mass decreases by one part in \(10^4\), the mass defect \(\Delta m\) is \(\Delta m = \frac{8.00 \text{ kg}}{10^4} = 0.0008 \text{ kg}\).

Step 2: Calculate the Energy Released Using Einstein's Equation

Utilize the mass-energy equivalence principle, \(E = \Delta m c^2\), where \(c\) is the speed of light in vacuum (\(c = 3 \times 10^8 \text{ m/s}\)). Thus, the energy released \(E\) is \[ E = 0.0008 \times (3 \times 10^8)^2 = 7.2 \times 10^{13} \text{ J}.\]

Step 3: Calculate the Average Power Developed

Power is defined as energy change per unit time. Given that the explosion occurs in \(4.00 \mu s = 4.00 \times 10^{-6} \text{ s}\), the average power \(P\) is calculated as follows:\[ P = \frac{7.2 \times 10^{13}}{4 \times 10^{-6}} = 1.8 \times 10^{19} \text{ W}.\]

Step 4: Calculate the Mass of Water Lifted

To find out how much water can be lifted to a height of 1.00 km, use the potential energy formula \(E = mgh\), where \(g = 9.81 \text{ m/s}^2\) is the acceleration due to gravity. Rearranging gives \(m = \frac{E}{gh}\).\[ m = \frac{7.2 \times 10^{13}}{9.81 \times 1000} = 7.34 \times 10^9 \text{ kg}.\]

Key Concepts

Mass-Energy Equivalence

At the heart of nuclear physics is the concept of mass-energy equivalence, a principle proposed by Albert Einstein and expressed in the famous equation \( E = mc^2 \). This equation tells us that mass can be converted into energy, and vice versa. Here, \( E \) represents energy, \( m \) is mass, and \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s. This powerful idea explains how small amounts of mass can be transformed into vast quantities of energy.

In nuclear reactions, such as explosions or fusion, some of the original mass seems to "disappear," becoming energy. This mass "loss" is called the mass defect, and it is precisely what fuels nuclear processes, making them incredibly potent. The mass-energy equivalence principle is fundamental to understanding how nuclear devices release energy and provides the groundwork for calculations involving atomic transformations.

Nuclear Explosion Energy Calculation

Nuclear explosions are fascinating and formidable because they release energy of staggering proportions. To calculate the energy released in an explosion, we start with the mass defect, which we identify based on the reduction in rest mass after the reaction. In our specific exercise, the original mass decreases by one part in \( 10^4 \). For 8 kg of plutonium, this results in a mass defect \( \Delta m = 0.0008 \) kg.

To convert this mass defect into energy, we use Einstein's equation \( E = \Delta m c^2 \). Substituting in the known values, the energy released in the explosion is \( 7.2 \times 10^{13} \) joules. Such a colossal amount of energy corresponds to the force released by nuclear reactions, highlighting just how a small change in mass results in extraordinary energy output.

Power Developed by Explosion

The concept of power is essential when discussing explosions because it describes how fast energy is released. Power is defined as the rate at which energy is transformed or transferred and is measured in watts (W). To determine the power developed by the nuclear explosion, we need the energy released and the duration over which this energy is unleashed.

In our scenario, the explosion occurs over \( 4.00 \mu s \) (microseconds), equivalent to \( 4.00 \times 10^{-6} \) seconds. By dividing the total energy released, \( 7.2 \times 10^{13} \) joules, by this time, we can calculate the power: \( 1.8 \times 10^{19} \) watts. This measurement of power illustrates how vastly different the time scales and rates are in nuclear events compared to typical chemical reactions or mechanical processes.

Potential Energy and Mass Calculations

Potential energy calculations allow us to interpret the impact that the released energy could have in terms of performing work, such as lifting a mass against gravity. Given the energy from a nuclear explosion, one interesting task is to determine how much mass can theoretically be lifted to a certain height. This involves using the potential energy formula \( E = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)), and \( h \) is the height.

For our exercise, we rearrange the formula to solve for mass: \( m = \frac{E}{gh} \). Plugging in the energy released from the explosion \( 7.2 \times 10^{13} \) joules, height 1 km, and \( g \), we find that \( 7.34 \times 10^9 \) kg of water could be lifted a kilometer into the air. This calculation underscores the tremendous capability of the energy in nuclear fuels to perform work, further illuminating the incredible power encompassed in nuclear phenomena.