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

Question

Steam is passed into 54 gm of water at 30° C till the temperature of mixture becomes 90° 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

Tardigrade Admin · Accepted Answer

Let mass of steam condensed =M gm M × 536 =m × 1 ×(100-90) =54 × 1 ×(90-30) M × 536 =54 × 60 M =(54 × 60/536) ≅ 6 gm Hence, mass of mixture =54+6=60 gm.

Q. Steam is passed into $54 g m$ of water at $3 0^{\circ} C$ till the temperature of mixture becomes $9 0^{\circ} C$ . If the latent heat of steam is $536 c a l / g m$ , the mass of the mixture will be

80 gm

60 gm

50 gm

24 gm

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

Q. Steam is passed into 54gm of water at 30∘C till the temperature of mixture becomes 90∘C. If the latent heat of steam is 536cal/gm, the mass of the mixture will be