Question

An iron ball of mass \[0.2kg\] is heated to $100^\circ C$. When it is put in an ice block at \[0^\circ C\] , 25g of ice is melted. The specific heat of iron in CGS units:
A) 1
B) \[0.1\]
C) \[0.8\]
D) \[0.08\]

Answer 1

Hint: We will use the concepts of latent heat and specific heat capacity to determine the specific heat of the iron. Since the iron ball is immersed in an ice block, the heat energy corresponding to the melting of ice must be equal to the drop in the temperature of the iron ball from $100^\circ C$ to \[0^\circ C\]
Formula used: In this solution, we will use the following formula:
$Q = mL$ where $Q$ is the amount of energy needed to convert the state of matter of a substance of mass $m$ and latent heat capacity $L$
$Q = mc\Delta T$ where $Q$ is the amount of energy needed to change the temperature of a substance of mass $m$ and specific heat capacity $c$ by temperature \[\Delta T\]

Complete step by step answer:
We’ve been given that an iron ball is placed in an ice block. We also know that in the process 25g of ice is melted.
Now the energy required to convert 25g of ice into the water must be equal to the energy required to cool the iron ball down to \[0^\circ C\] . This is because the ice will stop converting into the water only when the temperature of the iron ball will be \[0^\circ C\] . Hence comparing the specific heat of the ice ball and the latent heat energy of the ice block, we can write
${m_1}c\Delta T = {m_2}L$
Substituting the value of ${m_1} = 0.2\,kg = 200\,g$, $\Delta T = 100 - 0 = 100$ and ${m_2} = 25$ and $L = 80\,cal/g$, we get
$200 \times c \times 100 = 25 \times 80$
Solving for $c$, we get
$c = 0.1\,cal/g^\circ C$

Hence the correct choice is option (B).

Note: We must notice that the final temperature of the iron block will be \[0^\circ C\] since the constant interaction with the ice block will decrease its temperature. The conversion of ice into water will stop as soon as the iron ball reaches \[0^\circ C\] .