185g of steam at 100 degree Celsius is converted into water at 100 degree Celsius. How much heat ( in kilocalories ) is taken out of the system ? ( LHV of water is 540 cal/gm ) , if the answer is x, then value for x/20 is ?

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Solution

$A m o u n t o f h e a t = m \times L_{v} m = m a s s L_{v} = l a t e n t h e a t o f v a p o u r i s a t i o n . A m o u n t o f h e a t = 185 \times 540 = 99900 c a l / g S o x = 99900 c a l / g \frac{x}{20} = \frac{99900}{20} = 4995 c a l / g$

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185g of steam at 100 degree Celsius is converted into water at 100 degree Celsius. How much heat ( in kilocalories ) is taken out of the system ? ( LHV of water is 540 cal/gm ) , if the answer is x, then value for x/20 is ?

Amount of heat = m × Lvm = massLv = latent heat of vapourisation.Amount of heat = 185 × 540 = 99900 cal / gSo x= 99900 cal /gx20 = 9990020 = 4995 cal /g

$A m o u n t o f h e a t = m \times L_{v} m = m a s s L_{v} = l a t e n t h e a t o f v a p o u r i s a t i o n . A m o u n t o f h e a t = 185 \times 540 = 99900 c a l / g S o x = 99900 c a l / g \frac{x}{20} = \frac{99900}{20} = 4995 c a l / g$