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# 5 gm of steam at ${100^0}C$ passed into $6gm$ of ice at ${0^0}C$ . If the latent heats of steam and ice $n$ $cal$ $per$ $gm$ are $540$ and $80$ respectively, them the final temperature is:A) ${0^0}C$B) ${100^0}C$C) ${50^0}C$D) ${30^0}^{}C$

Last updated date: 20th Jun 2024
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Hint: To find the final temperature, the latent heat of steam and ice $540$ and $80$ respectively are given. First we need to find the heat of the ice melt and we need to find the total heat to ice at the given temperature then the temperature of the mixture can be calculated.

Formula used:
As we know that the latent heat is the amount of the substance and its specific latent heat is given by $mL$.
Where, $m$ is the mass of the substance
$L$ is the specific latent heat

Complete step by step solution:
Latent heat is energy absorbed or released by the body at the constant temperature
The amount of heat required to melt the ice will be $= 6 \times 80$ = $480cal$
The amount of heat required to raise the temperature of water by ${100^0}C$ $= 6 \times 1 \times 100$ = $600cal$
Then the total amount of heat required to raise the temperature of ice at ${100^0}C$ is $= 600 + 480$ = $1080cal$

To calculate the latent heat which is the product of the amount of the substance and its specific latent heat.
The amount of heat required to condense the steam $=$ $mL$ = $5 \times 540$ = $2700cal$
Then all the steam is not condensed.
The mass of the steam that is condensed is given by $=$ $m$
$\Rightarrow$ $m = \dfrac{{1080}}{{540}}$
$\Rightarrow$ $m = 2g$
Then the final temperature will be ${100^0}C$.

Therefore, the option (B) is the correct answer.

Note: The material or an element can change the phase, and change a state and also a large amount of heat changes occur without raising the temperature of the element or a material, then potentially a large loss of heat and in contact with phase changing or a heat absorbing material.