Heat And Thermodynamics Ques 7
- The temperature of $100 \mathrm{~g}$ of water is to be raised from $24^{\circ} \mathrm{C}$ to $90^{\circ} \mathrm{C}$ by adding steam to it. Calculate the mass of the steam required for this purpose.
(1996, 2M)
Show Answer
Answer:
Correct Answer: 7.( 12 g )
Solution:
Formula:
- Let $m$ be the mass of the steam required to raise the temperature of $100 \mathrm{~g}$ of water from $24^{\circ} \mathrm{C}$ to $90^{\circ} \mathrm{C}$.
Heat lost by steam $=$ Heat gained by water
$\therefore m\left(L+s \Delta \theta_1\right)=100 s \Delta \theta_2$ or $m=\frac{(100)(s)\left(\Delta \theta_2\right)}{L+s\left(\Delta \theta_1\right)}$
Here, $s=$ specific heat of water $=1 \mathrm{cal} / \mathrm{g}-{ }^{\circ} \mathrm{C}$,
$L=$ latent heat of vaporisation $=540 \mathrm{cal} / \mathrm{g}$.
$\Delta \theta_1=(100-90)=10^{\circ} \mathrm{C}$ and
$ \Delta \theta_2=(90-24)=66^{\circ} \mathrm{C} $
Substituting the values, we have
$ m=\frac{(100)(1)(66)}{(540)+(1)(10)}=12 \mathrm{~g} \Rightarrow \therefore m=12 \mathrm{~g} $
BETA
