Question

A lump of ice of $0.1kg$ at $- {10^ \circ }C$ is put in $0.15kg$ of water at ${20^ \circ }C$. If $\dfrac{x}{{100}}kg$ of ice is found in the mixture when it has reached thermal equilibrium. The value of $x\;$ is _______.
[Given : Specific heat of ice is $0.5kcalk{g^{ - 1}}{K^{ - 1}}$ and its latent heat of melting is$80kcal/kg$.]

Answer 1

In this question, we can calculate heat released by water and heat absorbed by ice by using the formula $\Delta Q = mc\Delta T$. We know that in the thermal equilibrium state, the balance heat can be used for the melting of ice (latent heat energy $\Delta Q = mL$). So, we can find the mass of melted ice.

Complete step by step answer:
According to the question, the mass of water is ${m_{water}} = 0.15kg$. The temperature of water is given ${20^ \circ }C$. We know that the specific heat of the water ${c_{water}} = 1kcalk{g^{ - 1}}{K^{ - 1}}$. The temperature of the water changes from ${20^ \circ }C$ to ${0^ \circ }C$, when the water is being cooled. So, heat released by the water to be cooled from ${20^ \circ }C$ to ${0^ \circ }C$ will be given as,
$\Delta {Q_{water}} = {m_{water}}{c_{water}}\Delta {T_{water}}$ …………….. (i)
Where $\Delta {T_{water}} = {20^ \circ }C - {0^ \circ }C = {20^ \circ }C$ is the change in temperature.
Putting the values of ${m_{water}},{\text{ }}{c_{water}}$ and $\Delta {T_{water}}$ in equation (i), we get
$\Delta {Q_{water}} = 0.15 \times 1 \times 20$
$\Rightarrow \Delta {Q_{water}} = 3kcal$
According to the question, we have given a lump of ice whose mass is ${m_{ice}} = 0.1kg$. We know that the specific heat of ice is ${c_{ice}} = 0.5kcalk{g^{ - 1}}{K^{ - 1}}$. The temperature changes from$- {10^ \circ }C$ to ${0^ \circ }C$. So, heat absorbed by the ice from$- {10^ \circ }C$ to ${0^ \circ }C$ will be given as-
$\Delta {Q_{ice}} = {m_{ice}}{c_{ice}}\Delta {T_{ice}}$ …………………………….(ii)
Where $\Delta {T_{ice}} = {0^ \circ }C - ( - {10^ \circ }C) = {10^ \circ }C$ is the change in temperature.
Putting the values of ${m_{ice}},{\text{ }}{c_{ice}}$ and $\Delta {T_{ice}}$ in equation (ii), we get-
$\Delta {Q_{ice}} = 0.1 \times 0.5 \times 10$
$\Rightarrow \Delta {Q_{ice}} = 0.5kcal$
So, Balance heat can be calculated as,
$\Delta Q = \Delta {Q_{water}} - \Delta {Q_{ice}}$
Putting the values of ${Q_{water}}$ and ${Q_{ice}}$ in the above equation, we get-
$\Delta Q = 3 - 0.5$
$\Rightarrow \Delta Q = 2.5kcal$ ……………………………….(iii)
Let $m{\text{ }}kg$ of ice be melted. We know that latent heat of ice $L = 80kcal/kg$. So, the thermal energy will be given as-
$\Delta Q = mL$
Putting the values of $m$ and $L$ in the above equation, we get-
$\Delta Q = m \times 80$ …………………………..(iv)
Now, we know that the balance heat can be used for melting ice.
So, comparing equation (iii) and (iv), we get-
$m \times 80 = 2.5$
$\Rightarrow m = \dfrac{{2.5}}{{80}}$
$\Rightarrow m = 0.03kg$
Thus, $0.03kg$ of ice is melted. So, $0.03kg$ is decreased from the mass of ice, and $0.03kg$ is increased from the mass of water.
So, the mass of ice which is left (after melting) is given as-
$m = 0.1 - 0.03$
$\Rightarrow m = 0.07kg$
Similarly, the mass of water which is left (after melting) is given as-
$m = 0.15 + 0.03$
$\Rightarrow m = 0.18kg$
The temperature of the ice and water mixture in the end (after melting of ice) is ${0^ \circ }C$.
According to the question, $\dfrac{x}{{100}}kg$ of ice is left in the mixture when it reaches thermal equilibrium.
So, $\dfrac{x}{{100}}kg = 0.07kg$
From the above equation, we can find the value of $x$.
$\dfrac{x}{{100}} = 0.07$
$x = 0.07 \times 100$
$x = 7$

Hence, the value of $x$ is $7$.

Note:
We know that specific heat is required heat to change the temperature of a substance while latent heat is the required heat for change of state. So, the balance heat is the difference of specific heat energy of water and ice which is equal to the latent heat energy of ice. The mass of melted ice is subtracted from the total mass of ice and added in the total mass of water. We have to find the value of $x$. So, we have to compare $\dfrac{x}{{100}}$ and $0.07$.