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Standard X

Physics

Calorimeter

1 g of steam ...

Question

$1 g$ 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 $= 80 c a l / g$

50^{o} C

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100^{o} C

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67^{o} C

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33^{o} C

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Solution

The correct option is B $100^{o} C$
Final temperature of mixture will be $100^{o} C$ . $1 g$ of steam releases $540 c a l$ of heat in changing its state into $1 g$ water of $100^{o} C$ . Out of $540 c a l$ , $80 c a l$ will be used by $1 g$ ice in changing its state into water of $0^{o} C$ . Now , remaining $460 c a l$ will raise the temperature of mixture upto $100^{o} C$ , as specific heat of water is $1 c a l / g -^{o} C$ .

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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

)

500 g

of water and

100 g

of ice at

0^{o} C

are in a calorimeter whose water equivalent is

40 g

10 g

of steam at

100^{o} C

added to it. Then water in the calorimeter is (Latent heat of ice

= 80 c a l / g

, Latent heat of steam

= 540 c a l / g

)

Q. How much heat is required to change

10 g

ice at

0^{o} C

to steam at

100^{o} C

? Latent heat of fusion and vaporization for

H_{2} O

are

80 c a l / g

and

540 c a l / g

respectively. Specific heat of water is

1 c a l / g

Q. 400 g of ice at 253 K is mixed with 0.05 kg of steam at

100^{o} C

. Latent heat of vaporisation of steam

= 540 c a l g^{- 1}

. Latent heat of fusion of ice

= 80 c a l g^{- 1}

. Specific heat of ice

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

. Find the resultant temperature of the mixture.

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

)

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1g of steam at 100oC and equal mass of ice at 0oC are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540cal/g, latent heat of ice =80cal/g

$1 g$ 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 $= 80 c a l / g$