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

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 \[m'=m\Rightarrow h=\frac{JL}{g}meter\]