5.0 Mechanical equivalent of heat

The water equivalent of a body is the amount of water that absorbs or given out the same amount of heat as is done by the body when heated or cooled through ${1^o}C$.

As the specific heat of water is unity, so the thermal capacity of a body ( $m \times s$ ) represents its water equivalent of the body.

So, the water equivalent is equal to the mass of the water ( as ${s_{water}} = 1$ ).

If the mechanical work is done $(W)$ on the body produces the same temperature change as the $H$ amount of heat given to the body.

Then we can write, $$W = JH$$ where $J$ is the mechanical equivalents of heat.

If $W$ and $H$ are measured in same units, then $J$ = 1 and it becomes a dimensionless quantity.

If $W$ is measured in Joule and $H$ is measured in calorie, then the unit of $J$ is joule per calorie.

The value of $J$ represents the number of joules of mechanical work is needed to raise the temperature of 1$g$ of water by ${1^o}C$.

Question 5. Calculate the amount of heat required to convert 1.00$kg$ of ice at ${-10^o}C$ into steam at ${100^o}C$ at normal pressure.

$\begin{equation} \begin{aligned} {s_{ice}} = 2100\,J\,k{g^{ - 1}}\,{K^{ - 1}}, \\ {L_{fusion}} = 3.36 \times {10^5}J\,k{g^{ - 1}}\,, \\ {s_{water}} = 4200J\,k{g^{ - 1}}\,{K^{ - 1}}, \\ {L_{vaporisation}} = 2.25 \times {10^6}J\,k{g^{ - 1}} \\\end{aligned} \end{equation} $

Heat neede to take the ice from ${-10^o}C$ to ${0^o}C$ = ${Q_1}$ $$\begin{equation} \begin{aligned} = \left( {1\,kg} \right)\left( {21000\,J\,k{g^{ - 1}}{K^{ - 1}}} \right)\left( {10\,K} \right) \\ = 2100\,J \\\end{aligned} \end{equation} $$Heat required to melt ice at ${0^o}C$ to water = ${Q_2}$ $$\begin{equation} \begin{aligned} = \left( {1\,kg} \right)\left( {3.36 \times {{10}^5}J\,k{g^{ - 1}}} \right) \\ = 336000\,J \\\end{aligned} \end{equation} $$Heat required to take 1$kg$ of water from ${0^o}C$ to ${100^o}C$ = ${Q_3}$$$\begin{equation} \begin{aligned} = \left( {1\,kg} \right)\left( {4200J\,k{g^{ - 1}}{K^{ - 1}}} \right)\left( {100\,K} \right) \\ = 420000\,J \\\end{aligned} \end{equation} $$Heat required to convert 1$kg$ of water at ${100^o}C$ into steam = ${Q_4}$$$\begin{equation} \begin{aligned} = \left( {1\,kg} \right)\left( {2.25 \times {{10}^6}J\,k{g^{ - 1}}} \right) \\ = 2250000\,J \\\end{aligned} \end{equation} $$Total heat required = $3.03 \times {10^6}\,J$.

Question 6. A 5$g$ of ice a ${-20^o}C$ is put into a 10$g$ of water at ${30^o}C$. Assuming that heat is exchanged only between the ice and the water, find the final temperature of the mixture.

$\begin{equation} \begin{aligned} {s_{ice}} = 2100\,J\,k{g^{ - 1}}\,{K^{ - 1}}, \\ {L_{fusion}} = 3.36 \times {10^5}J\,k{g^{ - 1}}\,, \\ {s_{water}} = 4200J\,k{g^{ - 1}}\,{K^{ - 1}} \\\end{aligned} \end{equation} $

The heat given by the water When it cools down from ${30^o}C$ to ${0^o}C$ is $$\begin{equation} \begin{aligned} = \left( {0.01\,kg} \right)\left( {4200\,J\,k{g^{ - 1}}{\,^o}{C^{ - 1}}} \right)\left( {{{30}^o}\,C} \right) \\ = 1260\,J \\\end{aligned} \end{equation} $$The heat required to bring the ice to ${0^o}C$ is$$\begin{equation} \begin{aligned} = \left( {0.005\,kg} \right)\left( {2100\,J\,k{g^{ - 1}}{\,^o}{C^{ - 1}}} \right)\left( {{{30}^o}\,C} \right) \\ = 210\,J \\\end{aligned} \end{equation} $$The heat required to melt the 5$g$of ice is$$\begin{equation} \begin{aligned} = \left( {0.005\,kg} \right)\left( {3.36 \times {{10}^5}\,J\,k{g^{ - 1}}\,} \right) \\ = 1680\,J \\\end{aligned} \end{equation} $$We can observe that the ice does not malt completely due to the required amount of heat is not supplied by the water. Some ice is only melted. So, the final temperature of the mixture is${0^o}C$

Question 7. A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is ${27^o}C$ and its melting point is ${327^o}C$.( ${L_{lead}} = 2.5 \times {10^4}\,J\,k{g^{ - 1}},{s_{lead}} = 125\,J\,k{g^{ - 1}}\,{K^{ - 1}}$ ).

Let take mass of bullet $m$

Heat required to take the bullet from ${27^o}C$ to ${327^o}C$$$\begin{equation} \begin{aligned} = m \times \left( {125\,J\,k{g^{ - 1}}\,{K^{ - 1}}} \right)\left( {300K} \right) \\ = m \times \left( {3.75 \times {{10}^4}\,J\,k{g^{ - 1}}} \right) \\\end{aligned} \end{equation} $$Heat required to melt the bullet=$$m \times \left( {2.5 \times {{10}^4}\,J\,k{g^{ - 1}}} \right)$$. If the initial speed be $v$, the kinetic energy is $\frac{1}{2}m{v^2}$ and hence the heat developed is $\frac{1}{2}\left( {\frac{1}{2}m{v^2}} \right) = \frac{1}{4}m{v^2}$. Thus, $$\begin{equation} \begin{aligned} \frac{1}{4}m{v^2} = m\left( {3.75 + 2.5} \right) \times {10^4}\,J\,k{g^{ - 1}} \\ v = 500\,m\,{s^{ - 1}} \\\end{aligned} \end{equation} $$.