Ratio of heat energy required to change in states of $100 g$ of water at $100^{\circ} C$ to steam to $540 g$ of ice to water respectively is $5 : y$ then $y$ = ___

Take latent heat of fusion $80 c a l / g m$ and latent heat of vaporization as $540 c a l / g m$

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Solution

Given

Mass of water $m_{1}$ = $100 g$

Mass of ice $m_{2}$ = $540 g$

Energy required to convert water to steam $Q_{1}$ = $100 \times 540$

Energy required to convert ice to water $Q_{2}$ = $540 \times 80$

$Q_{1}$ : $Q_{2}$ = $\frac{100 \times 540}{540 \times 80}$

= $10 : 8$ = $5 : 4$

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Similar questions

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

]

Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm