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Standard XII

Physics

Calorimetry

Ice at -20o...

Question

Ice at $- 20^{o}$ C is dropped into $0.25$ kg of the water at $20^{o}$ C. The final temperature of the system is $0^{o}$ C, when all the ice is melted. Find approximate mass of ice in the water (Specific heat of ice $= 2000 J k g^{- 1} K^{- 1}, L = 33.4 \times 10^{4} J k g^{- 1}$ ).

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Solution

Let the required amount of ice be $m K g$
Heat gain = Heat loss
$m \times 2000 \times [0 - (- 20)] + m \times 33.4 \times 10^{4} = 0.25 \times 4200 \times 20 \Rightarrow m = \frac{0.25 \times 4200 \times 20}{(2000 \times 20) + 33.4 \times 10^{4}} = 0.056 K g$

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2

kg of ice at

- 20^{o}

C is mixed with

5

kg of water at

20^{o}

C. The water content of the final mixture is (Latent heat of fusion

= 80

kcal/kg, specific heat of water

= 1

kcal/kg-

^{o} C

, specific heat of ice

= 0.5

kcal/kg-

^{o} C

200

g of ice at

0^{o}

C converts into water at

0^{o}

C in

1

minute when heat is supplied to it at a constant rate. In how much time

200

g of water at

0^{o}

C will change to

20^{o}

C? Take specific latent of ice

= 336 J g^{- 1}

Q. A piece of ice of mass

40 g

is dropped into

200 g

of water at

50^{o} C

. Calculate the final temperature of the water after all the ice has melted. Specific heat capacity of water

= 4200 J k g^{- 1} K^{- 1}

Specific latent heat of fusion of ice

= 336 \times 10^{3} J k g^{- 1}

A piece of ice of mass 40 g is dropped into 200 g of water at 50°C. Calculate the final temperature of water after all the ice has melted. Specific heat capacity of water = 4200 J/kg^oC , Specific latent heat of fusion of ice = 336 x 10³ J/kg.

Q. X g of ice at

0^{o}

C is added to

340

g of water at

20^{o}

C. The final temperature of the resultant mixture is

5^{o}

C. The value of X (in g) is closest to. [Heat of fusion of ice

= 333 J / g

; Specific heat of water

= 4.184 J / (g . K)

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Ice at −20oC is dropped into 0.25kg of the water at 20oC. The final temperature of the system is 0oC, when all the ice is melted. Find approximate mass of ice in the water (Specific heat of ice =2000Jkg−1K−1,L=33.4×104Jkg−1).

Let the required amount of ice be mKgHeat gain = Heat lossm×2000×[0−(−20)]+m×33.4×104=0.25×4200×20⇒m=0.25×4200×20(2000×20)+33.4×104=0.056Kg

Ice at $- 20^{o}$ C is dropped into $0.25$ kg of the water at $20^{o}$ C. The final temperature of the system is $0^{o}$ C, when all the ice is melted. Find approximate mass of ice in the water (Specific heat of ice $= 2000 J k g^{- 1} K^{- 1}, L = 33.4 \times 10^{4} J k g^{- 1}$ ).

Let the required amount of ice be $m K g$
Heat gain = Heat loss
$m \times 2000 \times [0 - (- 20)] + m \times 33.4 \times 10^{4} = 0.25 \times 4200 \times 20 \Rightarrow m = \frac{0.25 \times 4200 \times 20}{(2000 \times 20) + 33.4 \times 10^{4}} = 0.056 K g$