Question

Steam at ${100}^{\circ }C$ is passed into 20 g of water acquires a temperature of ${80}^{\circ }C$, the mass of water presents will be [Take specific heat of water 1 cal $g^{-1}{c}^{-1}$ latent heat of steam = $540g^{-1}$]

• A. $31.5g$
• B. $42.5g$
• C. $22.5g$
• D. $24g$

Answer 1

The correct option is C $22.5g$

Steam at ${100}^{o}C$, Water acquires= $20g$, Temperature= ${80}^{o}C$

The mass of water presents will be

Take specific heat of water= $1cal/g/c$

Latent heat of steam= $540g^{-1}$

Water ${80}^{o}C$ to ${100}^{o}C$, the heat=$mst$

$=20\times 1\times ({100}^o-{80}^o)$

$=20\times 20$

$=400cal$

Temperature ${100}^{o}C$ to ${100}^{o}C$ heat consumption= $mL$

$=20\times 540=10800$ cal

Total= $10800+400=11200cal$

$\Rightarrow \frac{11200}{400}g=22.5$

Mass of water will be= $22.5g$ .