Question

10 Grams of steam at ${100^\circ }C$ is mixed with 50 gm of ice at ${0^\circ }C$ then final temperature is-
(A) ${20^\circ }C$
(B) ${50^\circ }C$
(C) ${40^\circ }C$
(D) ${100^\circ }C$

Answer 1

We know temperature of ice will increase and temperature of steam will decrease. Both steam and ice will change their phase. By using latent heat of fusion and vaporization because phase is changing, we can solve this question. Heat that is used for ice and steam that will come into equilibrium. So, we will put both of them equal.

Complete step by step answer:
Temperature of ice will increase and the temperature of steam will decrease.
Given-
Mass of steam$\left( {{m_s}} \right) = 10gm$
Mass of Ice$\left( {{m_i}} \right) = 50gm$
Initial temperature of the steam${\left( {{T_i}} \right)_s} = {100^\circ }C$
Initial temperature of the Ice${\left( {{T_i}} \right)_I} = {0^\circ }C$
Now,
${\left( {Heat} \right)_{ice}} = {\left( {Heat} \right)_{Steam}}$
${\left( {mL} \right)_{ice}} + mS\left( {T - 0} \right) = {\left( {mL} \right)_{steam}} + mS\left( {100 - T} \right)$ -------$\left( 1 \right)$

Here L is the latent heat.
S is the specific heat of water,=$1C/gm$
Latent heat of fusion$= 80gm/cal$
Latent heat of evaporation$= 540gm/cal$
Now substitute all the values in equation $\left( 1 \right)$,
$4000 + 50T = 5400 + 10\left( {100 - T} \right)$
$\implies$ $4000 + 50T = 6400 - 10T$
$\implies$ $60T = 6400 - 4000$
$\implies$ $60T = 2400$
$\therefore$ $T = 40^\circ C$
So, final temperature is $T = 40^\circ C$

So, the correct answer is “Option C”.

Note:
Latent Heat is the heat or energy that is absorbed or released during a phase change of a substance. It can be either from liquid to solid or from a gas to liquid. Latent heat is related to a heat property which is known as enthalpy. There is a latent heat of fusion, Latent heat of vaporization. One more term, that is specific heat is the number of calories required to raise the temperature of $1gram$ of a substance by $1^\circ C$.