Question

A copper calorimeter of mass $100$ gm contains $200$ gm of a mixture of ice and water. Steam at ${100}^{\circ }C$ under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to ${50}^{\circ }C$ . If the mass of the calorimeter and its contents is now $330$ gm, what was the ratio of ice and water in the beginning? Neglect heat losses.
Given: Specific heat capacity of copper $=0.42\times {10}^{3}Jk{g}^{-1}{K}^{-1}$ ,
Specific heat capacity of water $=4.2\times {10}^{3}Jk{g}^{-1}{K}^{-1}$
Specific heat of fusion of ice $=3.36\times {10}^{5}Jk{g}^{-1}$
Latent heat of condensation of steam $$=22.5\times {10}^{5}Jk{g}^{-1}$$
A. $$1:1.26$$
B. $$1:5.26$$
C. $$5:1.26$$
D. $$7:1.86$$

Answer 1

Hint: When steam is passed, it will get converted to water. Find the mass of the steam converted to water. The steam from ${100}^{\circ }C$ cools to ${50}^{\circ }C$ , the energy given out during this temperature loss is responsible for the gain in temperature of the mixture.

Complete step by step answer:
If we read the question carefully, we will notice that the total mass is more than the initial mass. The final mass of the calorimeter and its contents is given to be $330$ gm. But the mass of the calorimeter and mixture of ice and water is $100$ gm and $200$ gm. So the amount of steam which was converted into water will be:
$$Totalmass-(massofcalorimeter+massofmixture)$$
$$\Rightarrow 330-(100+200)$$
$$\Rightarrow 30$$ gm.
Therefore, the mass of steam which got converted to water is $$30$$ grams.
The energy that is released from the steam is gained by the mixture of ice and water and by the calorimeter. It must be noted here that this process takes place in the following manner.
First steam from ${100}^{\circ }C$ is converted to water at ${100}^{\circ }C$.
Then the temperature of this water is reduced to ${50}^{\circ }C$ .
The energy gained from above process is utilised as follows:
To raise the temperature of calorimeter to ${50}^{\circ }C$
To convert Ice at $0^{\circ }C$ to water at $0^{\circ }C$ and then to increase the temperature of this water to ${50}^{\circ }C$
To increase the temperature of water in the mixture to ${50}^{\circ }C$ .
Let the mass of ice be ‘m’ kg. Then the mass of water initially will be ‘ $200-m$ ’ kg.
The mass of steam converted to water is $$30$$ grams.
The amount of heat required to change the temperature of water is:
$m_{0}S\mathrm{\Delta }T$
Where $m_0$ is the mass of water, $S$ is the specific heat capacity of water and $\mathrm{\Delta }T$ is the change in temperature.
Using this equation and the given values, we can have:
Energy lost to condense steam from ${100}^{\circ }C$ to water $=(0.1)(22.5\times {10}^5)$ --equation $1$
And the energy released in decreasing the temperature of water from ${100}^{\circ }C$ to ${50}^{\circ }C$ $=(0.03)(4.2\times {10}^3)(100-50)$ --equation $2$
Energy required to raise temperature of calorimeter to ${50}^{\circ }C$ $=(0.1)(0.42\times {10}^3)(50-0)$ --equation $3$
Energy required to convert ice at $0^{\circ }C$ to water $=(m)(3.36\times {10}^5)$ --equation $4$
Energy required to increase temperature of water from $0^{\circ }C$ to ${50}^{\circ }C$ $=(0.2-m)(4.2\times {10}^3)(50-0)$ --equation $5$

There is no heat loss, therefore, the energy lost from equation $1,2$ should be equal to energy gained in equation $3,4,5$
$$\Rightarrow (0.1)(22.5\times {10}^5)+(0.03)(4.2\times {10}^3)(100-50)$$
$=(0.1)(0.42\times {10}^3)(50-0)+(m)(3.36\times {10}^5)+(0.2-m)(4.2\times {10}^3)(50-0)$
Solving this we get:
$(0.18)(4.2\times {10}^3)(50)+m(3.36\times {10}^5)=(0.03)(22.5\times {10}^5)$
$\Rightarrow m=88.4gm$
Therefore, the mass of water $200gm-m=111.6gm$
Ratio of ice and water at the beginning $={\displaystyle \frac{88.4}{111.6}}$ which is approximately $={\displaystyle \frac{1}{1.26}}$

Hence, option A is the correct option.

Note: Don’t forget to convert the given mass into kilograms. Remember that specific heat of fusion of ice is the heat required to convert unit mass of ice to unit mass of water, without changing its temperature. And latent heat of condensation of steam is the amount of heat required to convert unit mass of steam to unit mass of water without changing its temperature.