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Standard XII

Physics

1st Law of Thermodynamics

10 gm of ice ...

Question

$10 g m$ of ice at $0^{o} C$ is kept in a calorimeter of water equivalent $10 g m$ . How much heat should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat)

6200 c a l

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7200 c a l

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13600 c a l

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8200 c a l

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Solution

The correct option is B $8200 c a l$
Heat supplied to ice to change from $0^{o} C$ ice to $100^{o} C$ vapour is
$M L_{f u s i o n} + M C Δ T + M L_{v a p}$
$Q_{1} = 10 \times 336 + 10 \times 4.2 \times 100 + 10 \times 2260$
$Q_{1} = 30160 J o r 7200 c a l$
Heat for calorimeter $Q_{2} = 10 \times 1 \times 100 = 1000 c a l$
$Q = Q_{1} + Q_{2} = 8200 c a l$

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Similar questions

10gm of ice at $- 20^{\circ}$ is dropped into a calorimeter containing 10gm of water at $10^{\circ} C$ , the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain.

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

)

250 g

of ice at

0^{\circ} C

converts into water at

0^{\circ} C

in a minute when a constant heat is supplied. In how much time (sec) the water thus obtained will change to

20^{\circ} C

? The specific latent heat of ice is

336 J g^{- 1}

. The specific heat capacity of water is

4.2 J g^{- 1}^{\circ} C^{- 1}

Q. A metallic calorimeter of mass 100g contains 200 g of ice and the initial temperature of the calorimeter and the ice is

- 10^{\circ} C

. Heat is supplied to the system containing calorimeter and ice at a constant rate of

50 c a l s^{- 1}

. Find the time required to raise the temperature of the system to

50^{\circ} C

. [Neglect the loss of heat to the surroundings].

Take specific heat capacity of calorimeter and ice as

0.2 c a l g^{- 1}^{\circ} C^{- 1}

and

\frac{1}{2} c a l g^{- 1}^{\circ} C^{- 1}

.Take latent heat of fusion of ice as

80 c a l g^{- 1}

and specific heat capacity of water as

1 c a l g^{- 1}^{\circ} C^{- 1}

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

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10gm of ice at 0oC is kept in a calorimeter of water equivalent 10gm. How much heat should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat)

The correct option is B 8200calHeat supplied to ice to change from 0oC ice to 100oC vapour isMLfusion+MCΔT+MLvapQ1=10×336+10×4.2×100+10×2260Q1=30160 J or 7200 calHeat for calorimeter Q2=10×1×100=1000 calQ=Q1+Q2=8200 cal

$10 g m$ of ice at $0^{o} C$ is kept in a calorimeter of water equivalent $10 g m$ . How much heat should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat)