   Chapter 29, Problem 83PE

Chapter
Section
Textbook Problem

Integrated ConceptsOne problem with x rays is that they are not sensed. Calculate the temperature increase of a researcher exposed in a few seconds to a nearly fatal accidental dose of x rays under the following conditions. The energy of the x-ray photons is 200 keV, and 4 .00×10 13 of them are absorbed per kilogram of tissue, the specific heat of which is 0.830   kcal / kg ⋅ ° C . (Note that medical diagnostic x-ray machines cannot produce an intensity this great.)

To determine

The temperature increase of a researcher exposed in a few seconds to a nearly fatal accident does of x-rays under the following conditions.

Explanation

Given Data:

Given that energy of the x-ray photon is200keV , and4.00×1013 of them are absorbed per kilogram of tissue, the specific heat of which is0.830kcal/kg.C

Formula Used:

Heat absorbed by tissue is calculated as

Q=NE

WhereQ= Heat absorbed

N= Number of protons

E= Energy of one photon

Below is the equation of heat to calculate the increase in temperature

Q=mCΔT

WhereQ= Heat absorbed

m= Mass

C= Specific Heat

ΔT= Change in temperature

Calculation:

We haveE=200eV

E=200eV×1.602×1019J1eV=3.204×1014J

andN=4.00×1013

Heat absorbed is calculated as follows:

Q=NE

Substituting the values in the above equation, we get

Q=(3.204×1014J)(4

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