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Standard XII

Physics

Calorimetry

Four cubes of...

Question

Four cubes of ice at -10 ^oC weighing one gram each is taken out from the refrigerator and are put in 150 grams of water at 20 ^oC. What is the final temperature of water when thermal equilibrium is attained? Assume that no heat is lost to the outside and water equivalent of container is 46 g. (Specific heat capacity of water = 1 cal/g-^oC, Specific heat capacity of ice = 0.5 cal/g-^oC, Latent heat of fusion of ice = 80 cal/g-^oC)

0 ^oC

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-10 ^oC

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17.9 ^oC

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None

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Solution

The correct option is C 17.9 ^oC
Heat gained by ice = Heat lost by water + Heat lost by container
Initial temperature of container = 20^oC
$4 \times \frac{1}{2} \times 10 + 4 \times 80 + 4 \times 1 \times (T - 0) = 196 \times 1 \times (20 - T) 20 + 320 + 4 T = 196 \times 20 - 196 T 200 T = 196 \times 20 - 340 T = \frac{3580}{200} = 17 . 9^{o} C$

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Similar questions

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

)

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

10 g

of ice at

- 20^{\circ} C

is added to

20 g

of water at

20^{\circ} C

. Assuming that no heat is lost to the surrounding, find the final temperature of the mixture.
[Take, specific heat capacity of ice

C_{i} = 2100 {J kg}^{- 1}^{\circ} C^{- 1}

, specific heat capacity of water

C_{w} = 4200 {J kg}^{- 1}^{\circ} C^{- 1}

& Latent heat of fusion of ice

L = 3 .36 \times 10^{5} {J kg}^{- 1}

]

Q. A 5g piece of ice at

- 20^{\circ}

C is put into 10g of water at

30^{\circ}

C. Assuming that heat is enchanged only between the ice and the water, find the final temperature of the mixture. Specific heat capacity of ice

= 2100 J k g^{- 1}^{\circ} C^{- 1}

. Specific heat capacity of water

= 4200 J k g^{- 1}^{\circ} C^{- 1}

and latent heat of fusion of ice

3.36 \times 10^{5} J k g^{- 1}

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

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The correct option is C 17.9 oCHeat gained by ice = Heat lost by water + Heat lost by container Initial temperature of container = 20o C 4×12×10 + 4×80 + 4×1×(T−0) = 196×1×(20−T)20+320+4T=196×20−196T200T=196×20−340T=3580200=17.9o C