Question

In a container of negligible mass $30$ g of steam at $100^\circ$C is added to $200$g of water that has a temperature of $100^\circ$C. If no heat is lost to the surroundings, what is the final temperature of the system? Also, find the masses of water and steam in equilibrium. Take $L_{v}=539 cal/g$ and $c_{water}= 1cal/g^\circ$C.

Answer 1

Hint: In the system, heat loss is zero. All the heat will remain in the system. So, the heat released by steam is gained by water to raise its temperature. Whenever there is a change in temperature, heat either will be released or gained. so, we assume the sum of the total heat is zero.
Formula used:
When there is change of temperature, we will use
 Q=mc$\Delta$T
Where, m is the mass, c is the specific heat and $\Delta$T is the temperature change.
When there is a change of state, we will use
$Q=m \times L$
Where, m is the converted mass from one state to another and L is the latent heat.

Complete answer:
In the question, given $ \textit{mass of steam}, m_{steam}= 30 g
 \textit {Temperature of steam }= 100 ^\circ C
  \textit{mass of water}, m_{water} =200 g
 \textit{ Temperature of water} = 40 ^\circ C$.
Assume Q is heat required to convert water ($200$ g) from $40 ^\circ$C to $100 ^\circ C $
So, we use the formula given by-
$Q = m_{water}c_{water} \Delta T $
Put the values of mass of water,
$Q=(200×1×(100-40)) = 12000 cal$
This Q heat is released by steam on converting into water. There is a change of state from steam to water so, we use the formula given by
$Q=m_{steam}L_{v}$ is the mass of steam that is converted into water.
$m_{steam}= \dfrac{Q}{L_{v}}$
Now put the of Q and $L_{v}$ in the above formula, we get
$m_{steam}= \dfrac{12000}{539} = 22.263 $ g
So, whole steam is not converted into water, so the final temperature of the system is equal to the temperature of steam.
$T_{final}=100^\circ C$
And at equilibrium, mass of water = $200+22.263 = 22.263$g.
mass of steam in system = $30-22.263 = 7.737$ g
So, after mixing temperature becomes $100^\circ C$ and masses of water and steam are $222.263$ g and $30-22.263 = 7.737$ g respectively.

Additional Information:
The significant conclusion of heat transfer is temperature change. If change in temperature is positive then heat is absorbed while if change in temperature is negative then heat will be released. So, heating raises the temperature while cooling drops it.

Note:
Practically, Some systems have lost their energy into surroundings but in the given question, it is given that heat loss is negligible, so heat released by steam is equal to heat absorbed by water to raise its temperature. Total heat lost in the surrounding is equal to zero.