Question

1 gm ice at $0^\circ C$ is mixed with 10 gm water at $50^\circ C$. The final temperature of mixture is.

Answer 1

When the given ice and water are mixed then, ice melts and its temperature increases while the temperature of water reduces. The final temperature is characterized by the equilibrium when the amount of heat energy absorbed by ice and produced water becomes equal to the heat energy lost by the hot water.

Formula used:
The heat energy released when ice melts is given in terms of mass and latent heat of fusion of ice by the following expression:
$Q = mL$
The amount of heat energy released or absorbed by a substance is given in terms of its mass, specific heat capacity and the change in temperature by the following expression.
$Q = mc\Delta T$

Complete step by step answer:
We are given 1 gram of ice which is at 0$^\circ$C. When it is mixed with 10 gm water at $50^\circ C$, then first the ice will undergo melting and it will absorb the amount of heat given by its latent heat of fusion. The value of latent heat of fusion of ice is 80 cal/g. Therefore, the amount of heat absorbed while melting is equal to $1g \times 80cal/g = 80cal$.
After melting, the water is formed which will further absorb more heat and its temperature will rise. Its initial temperature is 0$^\circ$C and let its final temperature be T. Also the specific heat capacity of water is equal to $1cal{\text{ }}{g^{ - 1}}^\circ {C^{ - 1}}$. Therefore, the amount of heat energy absorbed by this cold water is equal to $1g \times 1cal{\text{ }}{g^{ - 1}}^\circ {C^{ - 1}} \times \left( {T - 0} \right)$.
Now in the case of the hot water at 50$^\circ$C, it will lose its heat energy when it is mixed with ice at $0^\circ C$. The final temperature of the two liquids will be same i.e. T. Therefore, the amount of heat lost by the hot water is equal to $10g \times 1cal{\text{ }}{g^{ - 1}}^\circ {C^{ - 1}} \times \left( {50 - T} \right)$.
At the equilibrium, the amount of heat lost by hot water will be equal to the amount of heat gained by the ice. Therefore, we can equate the respective heat energies as follows:
$80cal + 1g \times 1cal{\text{ }}{g^{ - 1}}^\circ {C^{ - 1}} \times \left( {T - 0} \right) = 10g \times 1cal{\text{ }}{g^{ - 1}}^\circ {C^{ - 1}} \times \left( {50 - T} \right) \\\ \Rightarrow 80 + T = 10\left( {50 - T} \right) \\\ \Rightarrow 80 + T = 500 - 10T \\\ \Rightarrow 11T = 500 - 80 = 420 \\\ \Rightarrow T = \dfrac{{420}}{{11}} = 38.1^\circ C \\\$
This is the required value of final temperature.

Note:
It should be noted that the melting of ice takes place at the constant temperature of 0$^\circ C$ and the expression for heat energy absorbed while melting does not contain temperature as a parameter in it. All the heat energy absorbed or released at the melting or boiling point is used in changing the state of the substance rather than increasing or decreasing its temperature.