Question

In an energy recycling process, $\text{X} \, \text{g}$ of steam at $100 \,{}^\circ C$ becomes water at $100 \,{}^\circ C$ which converts $\text{Y} \, \text{g}$ of ice at $0 \,{}^\circ C$ into water at $100 \,{}^\circ C$ . The ratio of $\frac{X}{Y}$ will be (specific heat of water $=4200 \, J \, kg^{- 1} \, K$ , specific latent heat of fusion $=3.36\times 10^{5} \, J \, kg^{- 1}$ , specific latent heat of vaporization $=22.68\times 10^{5} \, J \, kg^{- 1}$ )

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $3$
• D. $2$

Answer 1

Answer: (A)

Heat loss = Heat gain
$\text{m}_{\text{1}} \text{L}_{\text{v}} = \text{m}_{\text{2}} \text{.L}_{\text{f}} + \text{m}_{\text{2}} \text{.S} \text{.} \Delta \text{T}$
$X\times 10^{- 3}\times 22.68\times 10^{5}$ $=Y\times 10^{- 3}\times 3.36\times 10^{5}+Y\times 10^{- 3}\times 4200\times 100$
$\therefore \frac{X}{Y}=\frac{7.56}{22.68}=\frac{1}{3}$