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Byju's Answer

Standard XII

Physics

Calorimetry

10 gm of ice ...

Question

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

10 gm ice and 10 gm of water

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20 gm of water

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5 gm ice and 15 gm of water

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20 gm ice

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Solution

The correct option is A 10 gm ice and 10 gm of water
As we know $Q = m c Δ θ$
$Q_{1} = 10 \times 1 \times 10 = 100 c a l$
$Q_{2} = 10 \times 0.5 (0 - (- 20)) + 10 \times 80 = (100 + 800) c a l = 900 c a l$
As $Q 1 < Q 2$ ,
so ice will not completely melt and final temeprature = $0^{\circ} C$
As heat given by water in cooling up to $0^{\circ} C$ is only just sufficient to increase the temperature of ice from $- 20^{\circ} C$ to $0^{\circ} C$
, hence mixture in equilibrium will consist of 10gm of ice and 10gm of water, both at 0∘C.

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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.

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

]

Q. The water equivalent of a calorimeter is

10

g and it contains

50

g of water at

15^{o}

C. Some amount of ice, initially at

- 10^{o}

C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice

= 0.5

cal

g m^{- 1}

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

, specific heat of water

= 1.0

cal

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

and latest heat of melting of ice

= 80

cal

g m^{- 1}

5 g m

of water at

30^{\circ} C

and

5 g m

of ice at

- 20^{\circ} C

are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice

= 0.5 c a l / g m^{\circ} C

and latent heat of ice

= 80 c a l / g m

10

g of ice at

- 20^{o}

C is dropped into a calorimeter containing

10

g of water at

10^{o}

C; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain.

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10 gm of ice at −20∘C is dropped into a calorimeter containing 10 gm of water at 10∘C. The specific heat of water is twice that of ice. Neglect heat capacity of the calorimeter. When equilibrium is reached, the calorimeter will contain :-

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