Question

At $30{}^\circ \text{ C}$ , a lead bullet 50 g, is fired vertically upwards with a speed of $840\text{ m/s}$ . The specific heat of lead is $0\cdot 02\text{ cal/g}{}^\circ \text{C}$ . On returning to the starting level, it strikes to a cake of ice at $0{}^\circ \text{C}$ . Calculate the amount of ice melted ( Assume all the energy is spent in melting only)
(A) $62\cdot 7\text{ g}$
(B) $55\text{ g}$
(C) $52\cdot 875\text{ kg}$
(D) $52\cdot 875\text{ g}$

Answer 1

Latent heat is defined as the heat or energy that is absorbed or released during a phase change of a substance. It could be either from a gas to a liquid or liquid to solid and vice-versa. Latent heat is related to a heat property called enthalpy
It is given by
$\text{L}=\dfrac{\text{Q}}{\text{m}}$
L=specific latent heat of a substance
m=mass
Q=energy released or absorbed
During phase change
Kinetic energy is given by
K.E $=\dfrac{1}{2}\text{ m}{{\text{v}}^{2}}$.

Complete step by step solution
Given,
$\begin{aligned} & \text{m}=50\text{ g}=\dfrac{50}{1000}\text{ kg}=\dfrac{1}{20} \\\ & \text{C}=0\cdot 02\text{ cal/g}{}^\circ \text{C} \\\ & {{\left( {{\text{T}}_{\text{i}}} \right)}_{\text{Bullet}}}=30{}^\circ \text{ C }{{\left( {{\text{T}}_{\text{i}}} \right)}_{\text{ice}}}=0{}^\circ \text{ C} \\\ & \text{v}=840\text{m/s} \\\ \end{aligned}$
Latent heat of ice $=80\text{cal/g}$
When a bullet is fired vertically upwards and returns to the starting level, only conservative force i.e. gravity acts on the bullet. That means mechanical energy is conserved.
i.e $\vartriangle$ K.E $+$ $\vartriangle$ P.E=0
$\vartriangle$ P.E=0
Because the final and initial positions are same
Then $\vartriangle$ K.E is also zero
Therefore, $\text{v}=840\text{m/s}$
$\because$ Initial K.E and final K.E are same
At this time, energy is given by
$\text{E}=\dfrac{1}{2}\text{ m}{{\text{v}}^{2}}$
Put all the value in this expression
$\begin{aligned} & \text{E}=\dfrac{1}{2}\left( \dfrac{1}{20} \right){{\left( 840 \right)}^{2}}\text{ J} \\\ & \text{=}\dfrac{1}{2}\times \dfrac{1}{20}\times 840\times 840\text{ J} \\\ & \text{=21}\times 840=17640\text{ J} \\\ \end{aligned}$
Convert this in call, then it becomes
$\begin{aligned} & \text{E}=\dfrac{176420}{4\cdot 2} \\\ & =4200\text{ cal} \\\ \end{aligned}$
This is the energy of bullet whole energy is spent in melting only
$\therefore$ Q=4200 cal
Now,
${{\left( {{\text{T}}_{\text{i}}} \right)}_{\text{Bullet}}}=30{}^\circ \text{C }{{\left( {{\text{T}}_{\text{i}}} \right)}_{\text{ice}}}=0{}^\circ \text{C}$
For melting of ice temperature of bullet should be equal to temperature of ice
$\therefore \text{ }\vartriangle \text{T}=30{}^\circ -0{}^\circ =30{}^\circ \text{C}$
And
${{\text{Q}}_{1}}=\text{mc}\vartriangle \text{T}$
i.e. heat spent to change the temperature of bullet to $0{}^\circ \text{C}$
$\begin{aligned} & \text{m}=50\text{g} \\\ & \text{C}=0\cdot 02\text{cal/}{{\text{g}}^{{}^\circ }}\text{C} \\\ & \vartriangle \text{T}=30{}^\circ \text{C} \\\ \end{aligned}$
Putting all values
$\begin{aligned} & {{\text{Q}}_{1}}=50\times 0\cdot 02\times 30 \\\ & \text{ }=30\text{cal} \\\ \end{aligned}$
Total heat given by bullet ${{\text{Q}}_{\text{T}}}=4200+30$
$=4230\text{cal}$
Now, entire heat of bullet is used in melting ice only,
Let M=mass of ice that method
L=latent heat of ice
Hence,
$\begin{aligned} & \text{m}\times \text{L}={{\text{Q}}_{\text{T}}} \\\ & 4230=\text{m}\times 80 \\\ & \text{m}=\dfrac{4230}{80} \\\ & \text{m}=52\cdot 88\text{g} \\\ \end{aligned}$
Option (D) is correct.

Note
Daily life is filled with examples of latent heat. Water has a high latent heat of fusion, so turning water into ice requires the removal of more energy than freezing liquid oxygen into solid oxygen, per unit gram. Latent heat plays a very important role in thunderstorms and hurricanes.