Question

In a calorimeter of water equivalent $20g$, water of mass $1.1\,kg$ is taken at $288K$ temperature. If steam at temperature $373K$ is passed through it and temperature of water increases by $6.5^\circ C$ then the mass of steam condensed is:
A) $17.5g$
B) $11.7g$
C) $15.7g$
D) $18.2g$

Answer 1

Hint:The heat gained by the water and calorimeter will be equal to heat lost from the steam as it is converted into liquid and then further cooled to bring its temperature to the final temperature of the system. Use formula for heat gained and lost with the above said context and solve to get the final answer.

Formula Used:
Heat lost by steam $ = xL + xc({T_1} - {T_2})$
Where, $x$ is the mass of steam condensed, $L = 540\,cal/g$ is the latent heat of fusion, $c = 1\,cal/g^\circ C$ is the specific heat capacity of steam, ${T_2}$ is the final temperature of the system, ${T_1}$ is the initial temperature of system.
Heat gained by water and calorimeter $ = (m + {m_c}) \times c \times ({T_2} - {T_1})$
Where, $m$ is the mass of water used, ${m_c}$ is the water equivalent of the calorimeter, $c = 1\,cal/g^\circ C$ is the specific heat capacity of steam, ${T_2}$ is the final temperature of the system, ${T_1}$ is the initial temperature of system.

Complete step by step solution:
First, make the units of all the given values similar.
Water equivalent of calorimeter, ${m_c} = 20g$
(we are not converting it into an SI unit. $c$ and $L$ are constants and use the cgs system for their units. If we convert this right now, we will have to convert it again later on to match with these constants or convert the value of $c$ and $L$ which will be a tedious task )
Mass of water, $m = 1.1kg = 1100g$
Initial temperature, ${T_1} = 288K = 15^\circ C$
Final temperature, ${T_2} = 15 + 6.5 = 21.5^\circ C$
Temperature of steam $ = 373K = 100^\circ C$
Now, when steam is passed to the calorimeter, its heat is absorbed by the calorimeter as well as the water. This same amount of heat which is absorbed by them is lost by steam and as a result of this heat loss, steam is first condensed and converted into liquid water and then its temperature is brought down to that of the calorimeter and the water in it.
These two heat values are equal and we must equate them to get our answer.
Heat lost by steam $ = xL + xc({T_1} - {T_2})$
Where, $x$ is the mass of steam condensed, $L = 540\,cal/g$ is the latent heat of fusion, $c = 1cal/g^\circ C$ is the specific heat capacity of steam, ${T_2}$ is the final temperature of system, ${T_1}$ is the initial temperature of system.
In this equation, we put values corresponding to the steam and get
Heat lost by steam $ = 540x + 1x(100 - 21.5)$
And, Heat gained by water and calorimeter $ = (m + {m_c}) \times c \times ({T_2} - {T_1})$
Where, $m$ is the mass of water used, ${m_c}$ is the water equivalent of the calorimeter, $c$ is the specific heat capacity of steam, ${T_2}$ is the final temperature of system, ${T_1}$ is the initial temperature of system.
In this equation, we put values corresponding to the steam and get
Heat gained by water and calorimeter $ = (1100 + 20) \times 1 \times (21.5 - 15)$
Equating these two heat values together, we get
$540x + 1x(100 - 21.5) = (1100 + 20) \times 1 \times (21.5 - 15)$
$ \Rightarrow 540x + 78.5x = 1120 \times 1 \times 6.5$
Which gives, $x(540 + 78.5) = 7280$
Therefore, $x = \dfrac{{7280}}{{618.5}} = 11.7g$

Therefore, Option B is the correct answer.

Note:Do not always convert values into SI units. Convert them accordingly as needed in the question as we did in this question. Otherwise it may lead to wrong answers or more tedious and hefty calculations. But make sure all the given values have similar units before actually solving the question. Don’t forget to include the units of constants when comparing units.