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Find the result of mixing ${\text{10}}\;{\text{gm}}$ ice at $ - 10^\circ {\text{C}}$ with ${\text{10}}\;{\text{g}}$water at ${\text{10^\circ C}}$. The specific heat capacity of ice is ${\text{2}}{\text{.1}}\;{\text{J}}{{\text{g}}^{ - 1}}{{\text{L}}^{ - 1}}$, Specific latent heat of ice is ${\text{336}}\;{\text{J}}{{\text{g}}^{ - 1}}$, and specific heat capacity of water is ${\text{4}}{\text{.2}}\;{\text{J}}{{\text{g}}^{ - 1}}{{\text{L}}^{ - 1}}$
Assumption:No loss of energy in the result of mixing of ${\text{10}}\;{\text{g}}$ ice at $ - 10^\circ {\text{C}}$ with of water at ${{10^\circ C}}$.

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Last updated date: 19th Jun 2024
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Answer
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Hint: In this question, first we need to understand the concept of latent heat. It states that the heat that is needed to convert a solid phase of an object into liquid phase or vapour phase, or, a liquid phase of an object into its vapour phase. Then find out the final temperature.
Complete step by step solution:

The transfer of heat will only occur when the ice and water both obtained ${{0^\circ C}}$, ice will gain heat and water will lose its latent heat, and the process will continue until both state achieves equal amount of heat, that is, the equilibrium occurs.
In the question the given quantities are the mass of the water is ${\text{10gm}}$, the specific heat of water is ${\text{4}}{\text{.2}}\;{\text{J}}{{\text{g}}^{{\text{ - 1}}}}{{\text{L}}^{{\text{ - 1}}}}$.
For water to come at $0^\circ {\text{C}}$, we calculate the amount of heat lost by water by using the formula,$H = m \times c \times \Delta T$
Now, we substitute the values in the above equation We get,
$ \Rightarrow H = 10 \times 4.2 \times \left( {0 - 10} \right)$
$\therefore H = 420\;{\text{J}}$
The heat released by water will be consumed by ice and ice will achieve ${\text{0}}^\circ {\text{C}}$, let’s denote the mass of ice ${m_1}$ and specific heat of ice ${c_1}$,
$H = {m_1} \times {c_1} \times \Delta T$
Now, we calculate the amount of heat absorbed by ice,
$H = 10 \times 2.1 \times \left( {0 - 10} \right)\;{\text{J}}$ [${{\text{m}}_{\text{1}}}{\text{ = 10gm}}$,${{\text{c}}_{\text{1}}}{\text{ = 2}}{\text{.1J}}{{\text{g}}^{{\text{ - 1}}}}{{\text{L}}^{{\text{ - 1}}}}$
$\therefore H = 210\;J$
Now both water and ice obtained ${{0^\circ C}}$, but water has some extra heat left in its phase. That extra amount of heat left is
$\Delta H = \left( {420 - 210} \right)$
$ \Rightarrow \Delta H = 210\;{\text{J}}$
So, taking the extra heat, some amount of ice will start melting.
We calculate the amount of the ice melt as
$\Delta H = {m_i} \times L$
Rearrange the equation as,
$m = \dfrac{{\Delta H}}{L}$
Now, substitute the values and we get,
${m_i} = \dfrac{{210}}{{336}}$
$ \Rightarrow {m_i} = 0.625\;{\text{g}}$
Now the equilibrium has been reached so no more ice will melt, neither the water will freeze. So the amount of ice that is left without getting converted into water phase: \[{m_1} = (10 - 0.625)\;{\text{g}} = 9.375\;{\text{g}}\] and amount of water in mixture left $m = (10 + 0.625) = 10.625\;{\text{g}}$ and the final temperature will be ${{0^\circ C}}$.

Note: As we know that the latent heat of fusion is the amount of heat required to melt the unit amount of the ice without change in the temperature of the system. The heat is used to change the phase of the ice.