Question

500 gram of water and 100 gram of ice at zero degrees Celsius are in a calorimeter whose water equivalent is 40 gram. 10 gram of steam at 100 degree Celsius is added to it. Then after equilibrium is reached, what is the mass of water in the calorimeter?
A. 580 gram
B. 590 gram
C. 600 gram
D. 610 gram

Answer 1

Hint: The water equivalent of a calorimeter is defined the amount of water needed to absorb the same amount of heat as that calorimeter does for one degree of increase in temperature.Use law of calorimetry to solve this type of questions which states that sum of all the heat gains and releases is zero.

Complete answer:
To convert a unit mass of substance from one state to another (viz. Solid, liquid gas), heat is required which is known as latent heat.
According to the law of calorimeters, the heat in an isolated system is conserved.
The water equivalent of a calorimeter is defined as the amount of water needed to absorb the same amount of heat as that calorimeter does for one degree of increase in temperature.
When steam reaches ${{0}^{{}^\circ }}C$, heat released by steam is
$\Delta {{Q}_{steam}}={{m}_{steam}}{{L}_{vap}}+{{m}_{steam}}{{s}_{water}}({{T}_{f}}-{{T}_{i}})$
$\Delta {{Q}_{steam}}=10\times 540+10\times 1\times (100-0)=6400cal$
Out of this 40 gram water equivalent or 40 calories are absorbed by calorimeter. Therefore, remaining heat which will convert ice to water is
$6400-40={{m}_{ice}}{{L}_{fusion}}={{m}_{ice}}\times 79.5$
$\Rightarrow {{m}_{ice}}=\dfrac{6360}{79.5}=80g$
This is the mass of ice which is converted to water. Mass of steam is also converted to water. Therefore, total mass of water in calorimeter is
$500+80+10=590g$

So, the correct answer is “Option B”.

Note:
The water equivalent of a calorimeter is defined as the amount of water needed to absorb the same amount of heat as that calorimeter does for one degree of increase in temperature.
The latent heat of fusion of water is 79.5 calories/g and latent heat of evaporation for water is 540calories/g.