Question

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

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

Answer 1

Answer: (D)

Steam at $100^{\circ }C$, Water acquires $=20g$,
Temperature $=80^{\circ }C$
The mass of water presents will be
Take specific heat of water $=1cal/g/c$
Latent heat of steam $=540g^{-1}$
Water $80^{\circ }C$ to $100^{\circ }C$, the heat $=mst$
$=20\times 1\times (100^{\circ }-80^{\circ }$ $=20\times 20$ $=400cal$ Temperature $100^{\circ }C$ to $100^{\circ }C$ heat consumption $=mL$
$<br/>=20\times 540=10800cal<br/>$
Total $=10800+400=11200cal$
$<br/>\Rightarrow \frac{11200}{400}g=22.5<br/>$
Mass of water will be $=22.5g$.