In an energy recycling process, textX textg of steam at 100 ° C becomes water at 100 ° C which converts textY textg of ice at 0 ° C into water at 100 ° C . The ratio of (X/Y) will be (specific heat of water =4200 J kg- 1 K , specific latent heat of fusion =3.36× 105 J kg- 1 , specific latent heat of vaporization =22.68× 105 J kg- 1 )

Question

In an energy recycling process, textX textg of steam at 100 ° C becomes water at 100 ° C which converts textY textg of ice at 0 ° C into water at 100 ° C . The ratio of (X/Y) will be (specific heat of water =4200 J kg- 1 K , specific latent heat of fusion =3.36× 105 J kg- 1 , specific latent heat of vaporization =22.68× 105 J kg- 1 ) (A) (1/3) (B) (2/3) (C) 3 (D) 2

Tardigrade Admin · Accepted Answer

Heat loss = Heat gain textm text1 textL textv = textm text2 text.L textf + textm text2 text.S text. Δ textT X× 10- 3× 22.68× 105 =Y× 10- 3× 3.36× 105+Y× 10- 3× 4200× 100 ∴ (X/Y)=(7.56/22.68)=(1/3)

$\frac{1}{3}$

$\frac{2}{3}$

$3$

$2$

Heat loss = Heat gain
$m_{1} L_{v} = m_{2} .L_{f} + m_{2} .S . Δ T$
$X \times 1 0^{- 3} \times 22.68 \times 1 0^{5}$ $= Y \times 1 0^{- 3} \times 3.36 \times 1 0^{5} + Y \times 1 0^{- 3} \times 4200 \times 100$
$∴ \frac{X}{Y} = \frac{7.56}{22.68} = \frac{1}{3}$

31​

32​

3

2

Heat loss = Heat gain m1​Lv​=m2​.Lf​+m2​.S.ΔT X×10−3×22.68×105 =Y×10−3×3.36×105+Y×10−3×4200×100 ∴YX​=22.687.56​=31​

$\frac{1}{3}$

$\frac{2}{3}$

$3$

$2$

Heat loss = Heat gain
$m_{1} L_{v} = m_{2} .L_{f} + m_{2} .S . Δ T$
$X \times 1 0^{- 3} \times 22.68 \times 1 0^{5}$ $= Y \times 1 0^{- 3} \times 3.36 \times 1 0^{5} + Y \times 1 0^{- 3} \times 4200 \times 100$
$∴ \frac{X}{Y} = \frac{7.56}{22.68} = \frac{1}{3}$