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An X-ray tube operates at 95 kV with a current of 25 mA and nearly all the electron energy goes...

Question:

An X-ray tube operates at 95 kV with a current of 25 mA and nearly all the electron energy goes into heat. If the specific heat of the 0.085-kg plate is 0.11 kcal/kg.°C, what will be the temperature rise per minute if no cooling water is used?

Electric Current

In the field of electro physics, current determines the quantity of passage of particles having some charge from one point to another in a circuit when a potential is generated. Current is quantified in the conventional units of ampere. This movement results in the electricity production.

Answer and Explanation:

Given data

  • The applied voltage to the x-ray tube is: {eq}{V_a} = 95\;{\rm{kV}} {/eq}.
  • The current flowing is: {eq}I = 25\;{\rm{mA}} {/eq}.
  • The mass of the plate is: {eq}M = 0.085\;{\rm{kg}} {/eq}.
  • The specific heat of plate is: {eq}c = 0.11\;{\rm{kcal/kg}}{\rm{.^\circ C}} {/eq}.


The expression for the power output is,

{eq}p = {V_a}I {/eq}


Substitute the values.

{eq}\begin{align*} p &= \left( {95\;{\rm{kV}}\left( {\dfrac{{1000\;{\rm{V}}}}{{1\;{\rm{kV}}}}} \right)} \right)\left( {25\;{\rm{mA}}\left( {\dfrac{{1\;{\rm{A}}}}{{1000\;{\rm{mA}}}}} \right)} \right)\\ &= 2375\;{\rm{W}} \end{align*} {/eq}


The expression for the temperature rise per minute when no cooling water is used is,

{eq}dT = \dfrac{P}{{Mc}} {/eq}


Substitute the values.

{eq}\begin{align*} dT &= \dfrac{{2375\;{\rm{W}}\left( {\dfrac{{1\;{\rm{J/s}}}}{{1\;{\rm{W}}}}} \right)}}{{\left( {0.085\;{\rm{kg}}} \right)\left( {0.11\;{\rm{kcal/kg}}{\rm{.^\circ C}}\left( {\dfrac{{4184\;{\rm{J}}}}{{1\;{\rm{kcal}}}}} \right)} \right)}}\\ &= 60.71\;{\rm{^\circ C/s}}\left( {\dfrac{{60\;{\rm{s}}}}{{1\;\min }}} \right)\\ &= 3642.6\;{\rm{^\circ C/min}} \end{align*} {/eq}


Thus, the temperature rise per minute when no cooling water is used is {eq}3642.6\;{\rm{^\circ C/min}} {/eq}.


Learn more about this topic:

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What is Electric Current? - Definition, Unit & Types

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7
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