Question

The $540\,g$ of ice at $0^\circ \,C$ is mixed with $540\,g$ of water at $80^\circ \,C$. The final temperature of the mixture is
A. $0^\circ \,C$
B. $40^\circ \,C$
C. $80^\circ \,C$
D. Less than $0^\circ \,C$

Answer 1

Hint:here, when the ice is mixed into water, the heat will be transferred from water to ice. This hat will melt the ice into the water. Therefore, the energy balanced equation will be Latent heat to convert ice at $0^\circ \,C$ into water at $0^\circ \,C$ $ + $ heating of water from $0^\circ \,C$ to $T^\circ \,C$ $ = $ cooling of hot water from $80^\circ \,C$ to $T^\circ \,C$.

Formula used:
Here, we will use the balanced equation to calculate the temperature of the mixture, which is given below
${m_1}H + {m_1}{c_1}\Delta T = {m_2}{c_2}\Delta T$
Here, ${m_1}$ is the mass of water, ${m_2}$ is the mass of ice, ${c_1}$ is the specific heat of water, ${c_2}$ is the specific heat of ice, $\Delta T$ is the change in temperature and $H$ is the latent heat of fusion.

Complete step by step answer:
Here, in the question, $$540\,g$$ of ice at $0^\circ \,C$ is mixed with $540\,g$ of water at $80^\circ \,C$. Therefore, latent heat of fusion or latent heat to convert ice into water, $H = 80^\circ C$
Now, the specific heat of water, ${c_1} = 1^\circ \,C$
Also, the specific heat of ice, ${c_2} = 0.5^\circ \,C$
As we know that when we mix the ice into water, the heat will transfer from water at $80^\circ \,C$ to the ice at $0^\circ \,C$ till the thermodynamic equilibrium will occur.
Since ice is saturated, therefore, it will absorb latent heat so that it can be converted into the water.

Let the final temperature of the mixture is, $ = \,T^\circ C$. Therefore, the energy balanced equation will be latent heat to convert ice at $0^\circ \,C$ into water at $0^\circ \,C$ $ + $ heating of water from $0^\circ \,C$ to $T^\circ \,C$ $ = $ cooling of hot water from $80^\circ \,C$ to $T^\circ \,C$

$ \Rightarrow \,{m_1}H + {m_1}{c_1}\Delta T = {m_2}{c_2}\Delta T$
$ \Rightarrow \,\left( {540 \times 80} \right) + \left( {540 \times 1 \times \left( {T - 0} \right)} \right) = \left( {540 \times 1 \times \left( {80 - T} \right)} \right)$
$ \Rightarrow 43200 + 540\,T = 43200 - 540\,T$
$ \Rightarrow \,540\,T = - 540\,T$
$ \Rightarrow \,1080\,T = 0$
$ \therefore \,T = 0$
Therefore, the temperature of the mixture will be $0^\circ \,C$.

Hence, option A is the correct option.

Note:Here, the specific heat of water and the specific heat of ice will be at room temperature. Also, the ice will be converted to water at room temperature. Also, we can say that in the above equation the heat lost by the water is equal to the heat gained by the ice. This is because the heat lost by the water is absorbed by the ice to get converted into the water.