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Standard X

Physics

Thermal Equilibrium

5 g of water ...

Question

$5 g$ of water at $30^{\circ} C$ and $5 g$ of ice at $- 20^{\circ} C$ are mixed together in a calorimeter. Find the final temperature of the mixture. Assume water equivalent of calorimeter to be negligible, specific heats of ice and water are $0.5$ and $1 c a l / g^{\circ} C$ , and latent heat of ice is $80 c a l / g$ .

0^{\circ} C

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10^{\circ} C

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- 30^{\circ} C

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> 10^{\circ} C

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Solution

The correct option is A $0^{\circ} C$
Let us consider that equilibrium is attained at $273 K .$
Then, heat absorbed by ice( $Q_{1}$ )=heat released by water( $Q_{2}$ )

For above condition, $5 \times .5 \times 20 = 50 < Q_{1} < 50 + 5 \times 80 = 450$

also, $5 \times 1 \times 30 = 150 < Q_{2} < 150 + 5 \times 80 = 550$

Since $Q_{1}$ and $Q_{2}$ can attain equality in the above ranges this is the correct assumption.

Hence, equilibrium temperature is $273 K$

So, the correct option is $A$

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Similar questions

Q. 5 g of water at

30^{\circ} C

and 5g of ice at

{- 20}^{\circ} C

are mixed together in a calorimeter. What is the final temperature of the mixture. Given specific heat of ice = 0.5 cal

g^{- 1}

^{\circ} C^{- 1}

and latent heat of fusion of ice = 80 cal

g^{- 1}

5

g of water at

30^{o}

C and

5

g of ice at

- 20^{o}

C are mixed in a calorimeter. Find the final temperature, water equivalent of calorimeter is negligible. Specific heat of ice is

0.5

cal

g^{- 1 o} C^{- 1}

and latent heat of ice is

80

cal per gram.

1 kg

of ice at

- 10^{\circ} C

is mixed with

4.4 kg

of water at

30^{\circ} C

. The final temperature of mixture is

Given specific heat of ice

= 0.5 {cal/g}^{\circ} C

,
Specific heat of water

= 1 {cal/g}^{\circ} C

,
Latent heat of fusion for ice

= 80 cal/g

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

50 g m

ice at -

10^{0} C

is mixed with

20 g m

steam at

100^{0} C .

When the mixture finally reaches its steady state inside a calorimeter of water equivalent

1.5 g m

then : [Assume calorimeter was initially at

0^{0} C,

Take latent heat of vaporization of water

= 540

cal/gm, Latent heat of fusion of water

= 80

cal/gm, specific heat capacity of water

= 1 c a l / g m -^{0} C,

specific heat capacity of ice

= {0.5 cal/ gm}^{0} C

]

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5g of water at 30∘C and 5g of ice at −20∘C are mixed together in a calorimeter. Find the final temperature of the mixture. Assume water equivalent of calorimeter to be negligible, specific heats of ice and water are 0.5 and 1cal/g∘C, and latent heat of ice is 80cal/g.