1g of steam is sent into 1g of ice. The resultant temperature of the mixture is:

(Latent heat of ice $L_{i c e} = 334 J$ , $c = 4.2 J$ , latent heat of vaporization $= 2258 J$ )

Question

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Answer 1

Energy required to convert 1g of ice at 0 $^{o}$ C to 100 $^{o}$ C water is

$= m L_{i c e} + m c Δ T = 1 [334 + 4.2 (100 - 0)] = 754 J$

Latent heat of vaporization of 1g steam $=$ $2260 J$

So, when a gram of steam is mixed with one gram of ice, ice converts to water at 100 $^{o}$ C and a part of steam converts to water at 100 $^{o}$ C.

The resultant mixture will be at 100 $^{o}$ C.

Answer 2

Energy required to convert 1g of ice at 0

^{o}

C to 100

^{o}

C water is

= m L_{i c e} + m c Δ T = 1 [334 + 4.2 (100 - 0)] = 754 J

Latent heat of vaporization of 1g steam

=

2260 J

So, when a gram of steam is mixed with one gram of ice, ice converts to water at 100

^{o}

C and a part of steam converts to water at 100

^{o}

C.

The resultant mixture will be at 100

^{o}

C.