Steam at 100° C is passed into 22g of water at 20° C . The mass of water that will be present when the water acquires a temperature of 90° C (Latent heat of steam is 540calg- 1 ) is

Question

Steam at 100° C is passed into 22g of water at 20° C . The mass of water that will be present when the water acquires a temperature of 90° C (Latent heat of steam is 540calg- 1 ) is (A) 24.8gm (B) 24gm (C) 36.6gm (D) 30gm

Tardigrade Admin · Accepted Answer

Heat loss = Heat gain mLV+mSW(100 - 90)=22Sw(90 - 20) m[.540+10].=22× 1× 70 m=(22 × 70/550)=2.8gm Mass of water =22+2.8=24.8gm

Q. Steam at $10 0^{\circ} C$ is passed into $22 g$ of water at $2 0^{\circ} C$ . The mass of water that will be present when the water acquires a temperature of $9 0^{\circ} C$ (Latent heat of steam is $540 c a l g^{- 1}$ ) is

$24.8 g m$

$24 g m$

$36.6 g m$

$30 g m$

Heat loss $=$ Heat gain
$m L_{V} + m S_{W} (100 - 90) = 22 S_{w} (90 - 20)$
$m [540 + 10] = 22 \times 1 \times 70$
$m = \frac{22 \times 70}{550} = 2.8 g m$
Mass of water $= 22 + 2.8 = 24.8 g m$

Q. Steam at 100∘C is passed into 22g of water at 20∘C . The mass of water that will be present when the water acquires a temperature of 90∘C (Latent heat of steam is 540calg−1 ) is

24.8gm

24gm

36.6gm

30gm