Question

A vessel of negligible heat capacity contains 40g of ice at ${{0}^{o}}C$. 8g of steam at ${{100}^{o}}C$ is passed into the ice to melt it. Find the final temperature of mixture (Latent heat of ice =$ 336$ $J{{g}^{-1}}$ ), Latent heat of steam = $2268$ $J{{g}^{-1}}$. ( Sp. Heat capacity of water = $4.2$ ${{g}^{-10}}{{C}^{-1}}$)

Answer 1

Hint: Apply the heat conservation law , Heat gain = Heat loss. Here heat taken by steam = heat taken by ice, ${{m}_{1}}{{L}_{1}}+{{m}_{1}}{{S}_{1}}{{t}_{1}}={{m}_{2}}{{L}_{2}}+{{m}_{2}}{{S}_{2}}{{L}_{2}}$ where 'm' is the mass , 'L' is the latent heat, 'S' is the specific heat capacity and 't' is change in temperature.

Complete step by step solution:
From your chemistry lessons you have learned about the heat law of conservation and it states that energy can neither be created nor be destroyed only it can convert from one form to another . So, the system will always have the same energy until some amount of energy is added from outside.
Therefore, Heat lost = Heat gain
Specific heat capacity is the energy that is needed to raise the temperature of 1 kg of substance by 1 k.
Latent heat capacity is the amount of heat required to convert the state of 1 kg of substance.
In the question the values that are given is,
mass of ice = $40g$
Mass of steam = $8g$
Specific heat capacity of water= $4.2$ $J{{g}^{-10}}{{C}^{-1}}$
Latent heat of ice = $336J{{g}^{-1}}$
Latent heat of steam = $2268J{{g}^{-1}}$
Initial temperature of both the ice and steam is gives as ${{0}^{o}}C\,and\,{{100}^{o}}C$
Let the take the final temperature of the mixture as ${{T}^{o}}C$ ,
 then heat taken by steam = heat taken by ice,
So, the formula is, ${{m}_{1}}{{L}_{1}}+{{m}_{1}}{{S}_{1}}{{t}_{1}}={{m}_{2}}{{L}_{2}}+{{m}_{2}}{{S}_{2}}{{L}_{2}}$……………… (1)
Where ${{m}_{1}}and\,{{m}_{2}}$ are the masses of steam and ice
${{L}_{1\,}}and\,{{L}_{2}}$ are the latent heat of steam and ice
${{S}_{1}}\,and\,{{S}_{2}}$ is the specific heat of water
 and ${{t}_{1\,}}and\,{{t}_{2}}$ is the change in temperature.
Now put all the values in equation one,
\[8\times 2268+8\times 4.2\times (100-T)=40\times 336+40\times 4.2\times (T-0)\]
\[18144+3360-33.6T=13400+168T\]
\[T=\dfrac{8064}{201.6}\]
\[\therefore T={{40}^{o}}C\]

Note: Specific heat of water will be equal to the specific heat of ice and steam. In latent heat the state changes but the temperature remains the same. Latent heat that is associated with melting or freezing is known as latent heat of fusion and which is associated with evaporation is known as latent heat of vapourization. In these cases latent heat of fusion is taking place.