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Calorimetry

5 g of steam ...

Question

$5 g$ of steam at $100^{\circ} C$ is passed into $6 g$ of ice at $0^{\circ} C$ . If the latent heat of steam and ice are $540 cal/g$ and $80 cal/g$ respectively and specific heat capacity of water is $1 cal / g^{\circ} C$ , then the final temperature of the mixture is

0^{\circ} C

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100^{\circ} C

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50^{\circ} C

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30^{\circ} C

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Solution

The correct option is B $100^{\circ} C$
Given:
Mass of steam, $m_{s} = 5 g$
Initial temperature of steam, $T_{s} = 100^{\circ} C$
Latent heat of steam, $L_{s} = 540 cal/g$
Mass of ice, $m_{i} = 6 g$
Initial temperature of ice, $T_{i} = 0^{\circ} C$
Latent heat of ice, $L_{i} = 80 cal/g$
Specific capacity of water, $s_{w} = 1 cal / g^{\circ} C$
To find:
Final temperature of mixture, $T = ?$

Heat required for ice to melt is $m L = 6 \times 80 = 480 cal$
Heat required to rise the temperature of water from $0^{\circ} C$ to $100^{\circ} C$ is $m s Δ T = 6 (1) (100) = 600 cal$

$∴$ Total heat required to convert ice at $0^{\circ} C$ to water at $100^{\circ} C$ is $1080 cal$

Let mass of steam condensed in this process be $m$
$m L_{v} = 1080 cal$
$m = 2 g$

Hence final mixture consists of $3 g$ of steam at $100^{\circ} C$ and $8 g$ of water at $100^{\circ} C$
Hence, final temperature of mixture is $100^{\circ} C$

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Similar questions

1 g

of steam at

100^{\circ} C

and an equal mass of ice at

0^{\circ} C

are mixed. The temperature of the mixture in steady state will be : (latent heat of steam

= 540 c a l / g

, latent heat of ice

= 80 c a l / g

, specific heat of water

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

)

60 g

of ice at

0^{0}

is mixed with

60 g

of steam at

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. At thermal equilibrium, the mixture contains:(Latent heats of steam and ice are

540 c a l g^{- 1}

and

80 c a l g^{- 1}

respectively, specific heat of water

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

)

Q. Ice at

0^{\circ} C

is added to

1 g

of steam at

100^{\circ} C

. For

x g

of ice added, the temperature of steam reduces to

0^{\circ} C

. Find the value of

x

.
(Latent heat of ice

= 80 cal/g

and latent heat of steam

= 540 cal/g

. Also, specific heat of water

= 1 cal/ kg K

)

1 g

of steam at

100^{o} C

and equal mass of ice at

0^{o} C

are mixed. The temperature of the mixture in steady state will be (latent heat of steam

= 540 c a l / g

, latent heat of ice

= 80 c a l / g

Q. The amount of heat required to convert

1 g

of ice at

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C is (specific heat of ice

= 0.5

cal/g/

^{o}

C, latent heat of ice

= 80

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5 g of steam at 100 ∘C is passed into 6 g of ice at 0 ∘C. If the latent heat of steam and ice are 540 cal/g and 80 cal/g respectively and specific heat capacity of water is 1 cal/g ∘C, then the final temperature of the mixture is

$5 g$ of steam at $100^{\circ} C$ is passed into $6 g$ of ice at $0^{\circ} C$ . If the latent heat of steam and ice are $540 cal/g$ and $80 cal/g$ respectively and specific heat capacity of water is $1 cal / g^{\circ} C$ , then the final temperature of the mixture is