Question

In the following equation calculate the value of $H$. 1kg steam at 200°C $ = $ H+ 1Kg water at 100°C (${S_{stream}}$=Constant$ = .5cal/gm^\circ C)$.

Answer 1

Hint:To calculate the total heat we add up the two heats released when steam at ${{200}^{0}}C$ gets converted to steam at ${{100}^{0}}C$. We can make use of the formula: Q= mcT and when there is change of state, we have to make use of the formula: Q= mL, where, m is the mass, c is the specific heat capacity of steam L is the latent heat of evaporation and T is the change in temperature.

Complete step by step answer:
We calculate the heat in two steps.
Step 1: Steam at 200°C loses out heat to change to steam at 100°C.Heat released in this case ${Q_1}= mcT$.
Where, m is the mass $\left( {m = 1000gm} \right)$, c is the specific heat capacity of steam $\left( {c = 0.5cal/gm^\circ C} \right)$ and T is the change in temperature $\left( {T = 100^\circ C} \right)$.
${Q_1} = 1000×0.5×100\\
\Rightarrow{Q_1}= 50000\,cal$

Step 2: Steam at 100℃ gets converted to water at 100℃. We know that the Latent heat of evaporation of water $ = $ 540cal/g℃.
${Q_2} = mL \\
\Rightarrow{Q_2}= 1000\times 540 \\
\Rightarrow{Q_2}= 540000\,cal$
So, the total heat released is,
$H = {Q_1}+{Q_2}\\
\Rightarrow H= 50000 + 540000 \\
\therefore H= 590000\,cal$

Hence, Value of $H$ is $590000\,cal$.

Note:Heat capacity or thermal capacity is a physical property of matter, defined as the amount of heat to be supplied to a given mass of a material to produce a unit change in its temperature.The enthalpy of vaporization, also known as the heat of vaporization or heat of evaporation, is the amount of energy that must be added to a liquid substance to transform a quantity of that substance into a gas.