Question

Steam is passed into $56\text{g}$ of water at ${30}^{\circ }\text{C}$ till the temperature of mixture becomes ${90}^{\circ }\text{C}$. If the latent heat of condensation of steam is $536\text{cal/g}$, the mass of the mixture will be:
(Specific heat capacity of water is $1{\text{cal/g}}^{\circ }\text{C}$)

• A. $80\text{g}$
• B. $62.1\text{g}$
• C. $60\text{g}$
• D. $54\text{g}$

Answer 1

The correct option is B $62.1\text{g}$

Let mass of steam condensed $=m$
Heat released $=$ Latent heat of condensation + $mc\left(\Delta T\right)$
where $c=$ Specific heat capacity of water
$\therefore$ Heat released $=(m\times 536+m\times 1\times ({100}^{\circ }-{90}^{\circ }\left)\right)\text{cal}$
As mixture is at ${90}^{\circ }\text{C}$ finally.
$\therefore$ Heat gained by water $=56\times c\times \Delta T$
$=56\times 1\times ({90}^{\circ }-{30}^{\circ })$
$=(56\times 1\times 60)\text{cal}$
We know,
Heat gain $=$ Heat loss
$\Rightarrow 56\times 1\times 60=m\times 536+m\times 1\times (100-90)$
$\Rightarrow m\times 546=56\times 60$
$\Rightarrow m=\frac{56\times 60}{546}\approx 6.1\text{g}$
$\therefore$ Mass of mixture $=(56+6.1)\text{g}\approx 62.1\text{g}$