Question

Steam is passed into $54\, gm$ of water at $30^{\circ} C$ till the temperature of mixture becomes $90^{\circ} C$. If the latent heat of steam is $536\, cal / gm$, the mass of the mixture will be

• A. 80 gm
• B. 60 gm
• C. 50 gm
• D. 24 gm

Answer 1

Answer: (B)

Let mass of steam condensed $=M\, gm$
$M \times 536 =m \times 1 \times(100-90)$
$=54 \times 1 \times(90-30)$
$M \times 536 =54 \times 60$
$M =\frac{54 \times 60}{536} \cong 6\, gm$
Hence, mass of mixture $=54+6=60\, gm.$