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Ratio and Proportion

How much stea...

Question

How much steam at $100^{o} C$ will just melt 2700 g of ice at $- 10^{o} C . (S p . h e a t o f i c e = 0.5 a n d l a t e n t h e a t o f s t e a m = 540 c a l g^{- 1})$

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Solution

Heat required to melt 2700 g of ice at
$- 10^{o} C = m \times S_{i c e} \times Δ t + m \times L_{i c e}$
$= [2700 \times 0.5 \times (0 - (- 10)] + (2700 \times 80]$
$= 229500 c a l$
Let M be the amount of steam required
$∴ H e a t r e q u i r e d = 229500 c a l .$
$ε_{0} \Rightarrow M \times 540 = 229500$
$M = \frac{229500}{540} = 425 g$

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Similar questions

100^{o}

2700

- 10^{o}

= 0.5

= 540

g^{- 1}

100^{\circ} C

- 10^{\circ} C

= 0.5 c a l g^{- 1}^{\circ} C^{- 1}

= 540 c a l g^{- 1},

= 80 c a l g^{- 1}

100^{o}

64

- 10^{o}

= 0.5

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= 540 c a l g^{- 1}

= 80 c a l g^{- 1}

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10 g

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