Question

Ice at $ - {20^ \circ }C$ is added to $50g$ of water at ${40^ \circ }C$ .When the temperature of the mixture reaches ${0^ \circ }C$ , it is found that $20g$ of ice is still unmelted. The amount of ice added to the water was close to (specific heat of water $4.2J/g{/^ \circ }C$ )
Specific heat of Ice $2.1J/g{/^ \circ }C$ Heat of fusion of water at ${0^ \circ }C = 334J/g$
A. $50g$
B. $40g$
C. $60g$
D. $100g$

Answer 1

Hint: To reach at the answer to the question, simply assume that the total mass of ice initially was $m\,g$ and out of that initial mass, only $20g$ of ice didn’t melt, i.e. it only used the heat given from the water to change its temperature from $ - {20^ \circ }C$ to ${0^ \circ }C$ . Now, since we have explained the situations, all you need to do is to develop the relation between the energy given by the water and the energy consumed by the ice to find the mass of the ice initially.

Complete step by step answer:
We will proceed with the solution exactly as explained in the hint section of the solution to the asked question. Firstly, we will assume an initial mass of $m\,g$out of which $20g$ of ice didn’t melt and only used the heat to get from $ - {20^ \circ }C$ to ${0^ \circ }C$ while, all the $50g$ of water gave up the heat to go from a temperature of ${40^ \circ }C$ to ${0^ \circ }C$ . This information will help us to easily develop the relation between the energy provided by the water and energy consumed by the ice.
Let us have a look at what is given to us in the question:
Mass of water, ${m_w} = 50g$
Initial temperature of water, ${T_w} = {40^ \circ }C$
Initial temperature of ice, ${T_i} = - {20^ \circ }C$
Specific heat of water, ${S_w} = 4.2J/g{/^ \circ }C$
Specific heat of ice, ${S_i} = 2.1J/g{/^ \circ }C$
Heat of fusion of water at ${0^ \circ }C = {H_f} = 334J/g$
Now, let us assume that the mass of ice initially was $ = {\kern 1pt} m\,g$
Out of this total initial mass, the mass which remained unmelted is = $20g$
Hence, the mass of ice that melted is $\left( {m - 20} \right)g$
Now, we can clearly see that all the energy is given by the water in the process, which can be given as:
Heat given by the water $ = {E_g} = {S_w} \times {m_w} \times \Delta {T_w}$
Since the final temperature of the whole system is ${0^ \circ }C$ , the value of temperature difference for water can be given as:
$\Delta {T_w} = {T_w} - {0^ \circ }C$
We know that the initial temperature of water was ${T_w} = {40^ \circ }C$
We can see that:
$
  \Delta {T_w} = {40^ \circ }C - {0^ \circ }C \\
  \Delta {T_w} = {40^ \circ }C \\
 $
Substituting the values of ${S_w},{\kern 1pt} \,{m_w}$ and $\Delta {T_w}$ , we get:
$
  {E_g} = 4.2 \times 50 \times 40 \\
  {E_g} = 8400 \\
 $
Now, we can write the heat received by the ice as:
${E_r} = {S_i} \times m \times \Delta {T_i} + {H_f}\left( {m - 20} \right)$
Upon substituting the values, we can write:
$
  {E_r} = 2.1 \times 20 \times m + 334\left( {m - 20} \right) \\
\Rightarrow {E_r} = 376m - 6680 \\
 $
We have already developed that:
${E_r} = {E_g}$
Substituting the values, we get:
$
  376m - 6680 = 8400 \\
\Rightarrow 376m = 15080 \\
\Rightarrow m = \dfrac{{15080}}{{376}} \\
 $
Upon solving, we get:
$m = 40.1g$
Hence, we can see that the value of mass is significantly closer to $40g$ as compared to all the other options. So, the answer is option (B).

Note:Many students get confused and believe that the system’s final temperature may not be ${0^ \circ }C$ since no information is given so. This is an important fact that until and unless all the ice has been converted into liquid, no heat will be used to increase the temperature and all the provided heat will be directed to convert the ice into liquid.