Question

$75g$ of water at ${100^0}C$ is added to $20g$ of ice at $- {15^0}C$. Determine the resulting temperature. Given that latent heat of ice $= 80ca{\lg ^{ - 1}}$ and specific heat of ice$= 0.5ca{\lg ^{ - 1}}{C^{ - 1}}$.

Answer 1

Hint We are given the temperature and the amount of water that is added to the ice at a given temperature. There will be a temperature change when the water is added to the ice. We are given the latent heat and the specific heat of ice. Now we have to find the resulting temperature after the water is added to the ice.

Complete Step by step solution
Let us take the final temperature of the mixture to be $T$.
Here, the water is at a higher temperature than ice. After the mixing water will lose some temperature and the temperature lost by the water will be gained by the ice. This temperature gained by the ice will be used by the ice to melt into water and then to raise the temperature of that water formed by the melting of ice.
Heat lost = heat gained
The specific heat of water is $1cal/g{/^ \circ }C$
Then the heat lost by water can be written as,
Heat lost $= ms\Delta T$
Where $m$ stands for the mass of the water, $s$ stands for the specific heat of water and $\Delta T$ stands for the change in temperature.
$m = 75g$,$s = 1cal/g/C$
$\Delta T = 100 - T$
Substituting,
Heat lost$= 75 \times 1 \times \left( {100 - T} \right)$
The heat gained by ice to change the temperature from $- {15^0}C$ to ${0^0}C$ will be,
Heat lost $= ms\Delta T$
Where $m$ stands for the mass of ice, $s$ stands for the specific heat of ice and $\Delta T$ stands for the change in temperature.
$m = 20g$,$s = 0.5cal/g/C$
$\Delta T = 0 - - 15 = 0 + 15$
Heat gain of ice $= 20 \times 0.5 \times \left( {0 + 15} \right) = 150cal$
At ${0^0}C$ the ice will start converting into water, this temperature can be given as,
Heat gain $= m \times L$
Where $m$ stands for the mass of ice and $L$ stands for the latent heat of ice.
Substituting $m = 20g$and $L = 80ca{\lg ^{ - 1}}$.
Heat gain $= 20 \times 80 = 1600cal$
The temperature of the water will rise from ${0^0}C$ to our final temperature $T$.
This can be written as,
Temperature change $= m \times s \times (T - 0)$
Substituting $m = 20g$ and the specific heat of water $L = 1cal/g{/^ \circ }C$
Temperature change $= 20 \times 1 \times T = 20T$
Heat lost = heat gained
Therefore,
$75 \times \left( {100 - T} \right) = 150 + 1600 + 20T$
Opening the bracket, we get
$7500 - 75T = 1750 + 20T$
Solving this equation, we get
$5750 = 95T$
From this,
$T = \dfrac{{5750}}{{95}} = {60.5^0}C$

The answer is : ${60.5^0}C$

Note
The amount of heat transferred per unit mass during the change of state of a substance is called Latent heat of the substance for the process. Latent heat of fusion is the quantity of heat required to convert $1kg$ of liquid to vapor at its boiling point.