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

Hint: 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{align}
 & \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{align} $
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{align}
  & \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{align} $
Convert this in call, then it becomes
 $ \begin{align}
  & \text{E}=\dfrac{176420}{4\cdot 2} \\
 & =4200\text{ cal} \\
\end{align} $
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{align}
  & \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{align} $
Putting all values
 $ \begin{align}
  & {{\text{Q}}_{1}}=50\times 0\cdot 02\times 30 \\
 & \text{ }=30\text{cal} \\
\end{align} $
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{align}
  & \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{align} $
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.