Question

Given, $6gm$ of steam at $100$℃ is mixed with $6gm$ of ice $0$℃. The mass of steam left uncondensed is
A. 2gm
B. 4gm
C. 3gm
D. 1gm

Answer 1

We need to go by steps to find the final answer. Firstly, loss of heat during condensation and then ice melting from both we can gain the mass of steam left uncondensed.
During solving we need to use the scientific actual values
${L_f} = 80cal/g$ ${L_f} = 80cal/g$ ${L_V} = 540cal/g$ ${L_V} = 540cal/g$ ${S_w} = 1cal/g -$ ${S_w} = 1cal/g -$℃
Formula used:
To solve above problem we will use following formulas listed below:
${Q_1} = m \times {L_v}$
Where Q is the heat
M is the mass of the steam
${L_v}$ is the latent heat of vapour
${Q_2} = {m_{ice}}{L_f} + {m_{ice}}{S_w}\Delta T$
Where M is the mass of the steam
∆T is the change in temperature so $0$℃ to $100$℃

Complete answer:
Since we know that, Condensation is the process where water vapour becomes liquid. It is the reverse of evaporation, where liquid water becomes a vapour.
Also, the heat required to convert a solid into a liquid or vapour, or a liquid into a vapour, without change of temperature is called latent heat
And the process of a solid becoming a liquid is called melting.
For steam,Heat lost by the steam during the condensation is calculated by:
${Q_1} = m \times {L_v} = 6 \times 540 = 3240cal$
Since the heat gained by the ice in melting and to rise its temperature from 0℃ to 100℃
So first here ice will melt and comes to liquid form as:
${Q_2} = {m_{ice}}{L_f} + {m_{ice}}{S_w}\Delta T$
{Q_2} = {m_{ice}}{L_f} + {m_{ice}}{S_w}\Delta T$$$$ = 6 \times 80 + 6 \times 1 \times 100 = 1080cal
So by this we can know the total steam did not condense into water.
Let 'm' gm of steam is condensed into water by giving $1080cal$
$m \times 540 = 1080$ $m \times 540 = 1080$ $m = \dfrac{{1080}}{{540}}$
$m = \dfrac{{1080}}{{540}}$
$m = 2gm$.

Therefore the correct option is A.

Note:
We need to go through the process it goes and how it affects the temperature and whether any created or not either heat is created or not whether their changes in mass during the transformation basically we need to go through their physical properties.