# JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Joule's Law (Heat and Mechanical Work)

Joule's Law (Heat and Mechanical Work)

Category : JEE Main & Advanced

Whenever heat is converted into mechanical work or mechanical work is converted into heat, then the ratio of work done to heat produced always remains constant. i.e. $W\propto Q$ or $\frac{W}{Q}=J$

This is Joule's law and J is called mechanical equivalent of heat.

(1) From W = JQ if Q = 1 then J = W. Hence the amount of work done necessary to produce unit amount of heat is defined as the mechanical equivalent of heat.

(2) J is neither a constant, nor a physical quantity rather it is a conversion factor which used to convert Joule or erg into calorie or kilo calories vice-versa.

(3) Value of  $J=4.2\,\frac{Joule}{cal}=4.2\times {{10}^{7}}\frac{erg}{cal}$

$=4.2\times {{10}^{3}}\frac{Joule}{kcal}$.

(4) When water in a stream falls from height h, then its potential energy is converted into heat and temperature of water rises slightly.

From     $W=JQ$ $\Rightarrow$  mgh = J (mc $\Delta \theta$)

[where m = Mass of water, c = Specific heat of water, $\Delta \theta =$ temperature rise]

$\Rightarrow$ Rise in temperature $\Delta \theta =\frac{gh}{Jc}{}^\circ C$

(5) The kinetic energy of a bullet fired from a gun gets converted into heat on striking the target. By this heat the temperature of bullet increases by$\Delta \theta$.

From     W = JQ    $\Rightarrow$ $\frac{1}{2}m{{v}^{2}}=J(\,m\,s\,\Delta \theta )$

[where m = Mass of the bullet, v = Velocity of the bullet,     c = Specific heat of the bullet]

$\Rightarrow$ Rise in temperature $\Delta t=\frac{{{v}^{2}}}{2Jc}{}^\circ C$

If the temperature of bullet rises upto the melting point of the bullet and bullet melts then.

From     $W=J({{Q}_{Temperature\text{ }change}}+{{Q}_{Phase\text{ }change}})$

$\Rightarrow$ $\frac{1}{2}m{{v}^{2}}=J(mc\,\Delta \theta +mL)$;    L = Latent heat of bullet

$\Rightarrow$  Rise in temperature   $\Delta \theta =\left[ \frac{\left( \frac{{{v}^{2}}}{2J}-L \right)}{c} \right]\,{}^\circ C$

(6) If m kg ice-block falls down through some height (h) and melts partially (m' kg) then its potential energy gets converted into heat of melting.

From  W = JQ  $\Rightarrow$ $mgh=J\,m'L$ $\Rightarrow$ $h=\frac{m'}{m}\left( \frac{JL}{g} \right)$

If ice-block melts completely then $m'=m\Rightarrow h=\frac{JL}{g}meter$

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