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Steam at ${100^ \circ }C$is passed into 1.1kg of water contained in a calorimeter of water equivalent 0.02kg at ${15^ \circ }C$ till the temperature of the calorimeter and its contents rises to ${80^ \circ }C$. The mass of the steam condensed in kg is
A. 0.130
B. 0.065
C. 0.260
D. 0.135

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint:Heat or thermal energy is the form this energy possesses when it is being transferred between systems and surroundings, this flow of energy is referred to as heat. To solve the given question, we will first list the data needed and then apply the formula of heat developed in a body and equate it with power.
Formula Used:
$Q=mc\Delta T$

Complete step-by-step solution:
According to the question,
Mass of the water contained in a calorimeter \[=\text{ }\left( 1.10\text{ }+\text{ }0.02 \right)\text{ }=\text{ }1.12kg\]
Specific heat capacity of the water $=4.184\times {{10}^{3}}J/kgK$
Now, as we know when we supply heat to a body, its temperature increases. The amount of heat absorbed by the body depends upon its mass, the change in body depends upon the material of the body as well as the surrounding conditions, such as pressure.
Mathematically, we write the equation as
$Q=mc\Delta T$$--(i)$
Where ‘$\Delta T$’ is a change in temperature, ‘m’ is the mass of the body and ‘Q’ represents the heat supplied, ‘c’ represents the specific heat capacity of the substance.
So, from the above relation heat gained by the calorimeter and water is given as,
$Q=1.12\times 4.18\times {{10}^{3}}\times 65\cdots \cdots \left( 1 \right)$
Let us assume that the mass of the steam is ‘m’ kg.
Then, latent heat of vaporization of water is given as,
Latent heat of vaporization of water at ${{100}^{\circ }}C$ is given as:
$540cal/g=540\times 4.184\times {{10}^{3}}J/kg$
Heat lost by the steam is given by,
${{Q}^{'}}=m\left( 540\times 4.184\times {{10}^{3}} \right)+m\left( 4.184\times {{10}^{3}}\times \left( 100-80 \right) \right)J/kgK\cdots \cdots \left( 2 \right)$
Now, clearly the heat lost by the steam must be equal to the heat gained by the calorimeter and water. So we should equate equations (1) and (2). Thus we have:
$\begin{align}
  & m\times 4.184\times {{10}^{3}}\times \left( 540+20 \right)=1.12\times 4.18\times {{10}^{3}}\times 65 \\
 & \Rightarrow m\left( 540+20 \right)=1.12\times 65 \\
 & \Rightarrow m=\dfrac{1.12\times 65}{560} \\
 & \therefore m=0.13kg \\
\end{align}$
Thus, the mass of the steam condensed in kg is 0.13.
Hence, option (A) is the correct answer.

Note: In order to solve these kinds of questions students must be aware of what a calorimeter is, and what its function is. A calorimeter is a device which is used to measure the heat produced during a chemical, mechanical or electrical reaction. It is also used to calculate the heat capacity of various materials.


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