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Standard XI

Physics

Calorimetry

10 gm of ice ...

Question

$10 gm$ of ice cubes at $0^{\circ} C$ are released in a tumbler (water equivalent $55 gm$ ) at $40^{\circ} C$ . Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly ( $L = 80$ $cal/g$ )

{31.54}^{\circ} C

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{21.54}^{\circ} C

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19^{\circ} C

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{15.68}^{\circ} C

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Solution

The correct option is B ${21.54}^{\circ} C$
Let the final temperature be $T$
Heat gained by ice is heat required to melt ice to water at $0^{\circ} C$ and heat required to increase temperature of water from $0^{\circ} C$ to $T^{\circ} C$
$= m L + m \times s \times (T - 0)$
$= 10 \times 80 + 10 \times 1 \times T$
Heat lost by tumbler $= 55 \times 1 \times (40 - T)$
By using law of calorimetry,
$800 + 10 T = 55 \times (40 - T)$
$\Rightarrow T = {21.54}^{\circ} C$

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Similar questions

10 gm

of ice cubes at

0^{\circ} C

are released in a tumbler (water equivalent

55 gm

) at

40^{\circ} C

. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (

L = 80

cal/g

)

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

Q. Four cubes of ice at

- 10^{\circ} C

, each of one gram, is taken out from a refrigerator and dropped into

150 g

of water at

20^{\circ} C

. Find the temperature of the water when thermal equilibrium is attained. Assume that no heat is lost to the outside and the water equivalent of the container is

46 g

. (Specific heat capacity of water

= 1 cal/g^{\circ} C

, specific heat capacity of ice

= 0.5 cal/g^{\circ} C

, latent heat of fusion of ice

= 80 cal/g

)

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. Two

50 g m

ice cubes are dropped into

250 g m

of water into a glass. If the water was initially at a temperature of

25^{\circ} C

and the temperature of ice

- 15^{\circ} C

. Find the final temperature of water(in Celcius). (specific heat of ice

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

and

L = 80 c a l / g m

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10 gm of ice cubes at 0∘C are released in a tumbler (water equivalent 55 gm) at 40∘C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/g)

$10 gm$ of ice cubes at $0^{\circ} C$ are released in a tumbler (water equivalent $55 gm$ ) at $40^{\circ} C$ . Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly ( $L = 80$ $cal/g$ )