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 $=540 g ^{-1}$ ]

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