Question

If 10 grams of ice and 1 gram of steam are allowed to attain thermal equilibrium, what will be the resultant temperature?

Answer 1

Hint: Use principle of calorimetry, which states that, in an isolated system the total heat lost by a hot body is equal to the total heat gained by cold body. This law is based on the law of conservation of energy.
Calculate expressions for heat lost by steam and heat gained by ice to reach an equilibrium temperature and then equate these expressions.

Complete step-by-step answer:
According to the principle of calorimetry, the sum of all the heat gains and heat lost in an isolated system is zero. Heat transfer continues until the system is in thermal equilibrium.
Therefore, heat lost by steam will be equal to heat gained by ice.
$\Delta {{Q}_{steam}}+\Delta {{Q}_{ice}}=0$
We are given that,
${{m}_{ice}}=10g$
${{m}_{steam}}=1g$
We assume that the equilibrium temperature is $T$.
We first calculate heat gained by ice to convert into water at ${{0}^{{}^\circ }}C$.
$\Delta {{Q}_{1}}={{m}_{ice}}{{L}_{fusion}}$
$\Delta {{Q}_{1}}=10\times 80\,calories=800cal$
(Latent heat of fusion of water at ${{0}^{{}^\circ }}C$ is $80cal/g$)
Heat gained by mass of ice, converted into water, to reach equilibrium,
$\Delta {{Q}_{2}}={{m}_{ice}}{{s}_{water}}(T-0)$
$\Delta {{Q}_{2}}=10\times 1\times T=10T$
(specific heat of water is$1\dfrac{cal}{{{g}^{{}^\circ }}C}$)
Heat lost by steam when it converts into water at ${{100}^{{}^\circ }}C$.
$\Delta {{Q}_{3}}={{m}_{steam}}(-{{L}_{vapourisation}})$
$\Delta {{Q}_{3}}=1\times (-536)cal=-536cal$
(Latent heat of condensation is negative of latent heat of vaporization)
Heat lost by mass of steam converted into water to reach thermal equilibrium is,
$\Delta {{Q}_{4}}={{m}_{steam}}{{s}_{water}}(T-100)$
$\Delta {{Q}_{4}}=1\times 1\times (T-100)=T-100$
According to principle of calorimetry,
$\Delta {{Q}_{1}}+\Delta {{Q}_{2}}+\Delta {{Q}_{3}}+\Delta {{Q}_{4}}=0$
$\Rightarrow 800+10T+(-536)+(T-100)=0$
Solving this, we get
$T=-{{14.9}^{{}^\circ }}C$
Since the equilibrium temperature must lie between ${{T}_{ice}}$ and ${{T}_{steam}}$, we are not getting a physical result. This shows that all the ice will not melt and therefore $T={{0}^{{}^\circ }}C$.

Note: The equilibrium temperature of a mixture always lies between temperature of hot and cold substance. We can also do this kind of problem by checking the amount of heat released and absorbed when converting into a different state and then find the state of the mixture and the temperature of the mixture.