Question

A 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 rises to $80^\circ C$. What is the mass of the steam condensed?
Latent heat of steam $ = 536cal/g$
(A) $0.31kg$
(B) $0.13kg$
(C) $0.41kg$
(D) $0.15kg$

Answer 1

Hint
The amount of heat that is gained by the water and calorimeter is also the amount of heat that is by the steam. So we can substitute the values given in the question and then equate the two equations, and from there calculate the mass of steam.
In this solution we will be using the following formulas,
$\Rightarrow \Delta Q = mc\Delta T$
where $\Delta Q$ is the amount of heat lost or gained due to the change in temperature $\Delta T$
$m$ is the mass and $c$ is the specific heat capacity.
And $\Delta Q = ml$ where $\Delta Q$ is the amount of heat lost or gained due to the change in state and $l$ is the latent heat.

Complete step by step answer
The heat that is lost by the steam in converting its state to water and then thereby reducing its temperature to $80^\circ C$, is the heat that is gained by the water and calorimeter to rise to the temperature of $80^\circ C$.
Therefore, we can calculate the heat that is lost by steam as,
$\Rightarrow \Delta Q = {m_s}l + {m_s}c\Delta T$
Now after the steam is converted to water at $100^\circ C$ and the latent heat of steam is given as $l = 536cal/g$. Then the specific heat capacity of water has a value of $c = 1Cal/kg^\circ C$ for the change in temperature from $100^\circ C$ to $80^\circ C$. Therefore by substituting the values we get,
$\Rightarrow \Delta Q = \left( {{m_s} \times 536} \right) + \left[ {{m_s} \times 1 \times \left( {100 - 80} \right)} \right]$
By taking ${m_s}$ common from both the terms, we have,
$\Rightarrow \Delta Q = {m_s}\left[ {536 + 20} \right]$
This, on calculating gives us,
$\Rightarrow \Delta Q = 556 \times {m_s}$
Now for the calorimeter, it contains water of mass $1.1kg$ and the calorimeter itself is water equivalent of $0.02kg$. So we can take the total mass of water as, $\left( {1.1 + 0.02} \right)kg$, which is equal to ${m_w} = 1.12kg$.
The specific heat of water as mentioned earlier is $c = 1Cal/kg^\circ C$ and the temperature in this case changes from $15^\circ C$ to $80^\circ C$. So the change in temperature is $\Rightarrow \left( {80 - 15} \right)^\circ C$
Therefore, the amount of heat gained in this case is given by,
$\Rightarrow \Delta Q = {m_w}c\Delta T$
Substituting the values we get,
$\Rightarrow \Delta Q = 1.12 \times 1 \times \left( {80 - 15} \right)$
On doing the calculation we get the heat gained as,
$\Rightarrow \Delta Q = 72.8Cal$
Now the heat gained in this case is equal to the heat lost by the steam, so we can write,
$\Rightarrow 72.8 = 556 \times {m_s}$
Therefore, from here we can find the value of ${m_s}$ as,
$\Rightarrow {m_s} = \dfrac{{72.8}}{{556}}kg$
This gives us the mass of steam as,
$\Rightarrow {m_s} = 0.13kg$
So the correct answer is option (B); $0.13kg$.

Note
The specific heat of a substance is the amount of heat that is required to raise the temperature of unit mass of that substance by an amount of $1^\circ C$. And the latent heat capacity of a substance is the amount of heat required for unit mass of that substance to undergo phase transition.