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Standard XII

Physics

Calorimetry

Ice at 20 C...

Question

Ice at $20 C$ is added $t o 50 g$ of water at $40 C$ .When the temperature of the mixture reaches $0 C$ ,it is found that $20 g$ of ice is still unmelted. The amount of ice added to the water was close to (specific heat of water $= 4.2 J / g /^{o} C)$
Specific heat of Ice $= 2.1 J / g /^{o} C$ Heat of fusion of water at $0^{o} C = 334 j / g$

50 g

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40 g

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60 g

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100 g

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Solution

The correct option is D $40 g$
Let the amount of ice is m gm.
According to the principal of calorimeter
heat taken by ice = heat given by water
$∴ 20 \times 2.1 \times m + (m - 20) \times 334$
$= 50 \times 4.2 \times 40$
$376 m = 8400 + 6680$
$m = 40.1$

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Similar questions

Q. Ice at

- 20^{\circ} C

is added to

50 g

of water at

40^{\circ} C

, when the temperature of the mixture reaches

0^{\circ} C

, it is found that

20 g

of ice is still unmelted. The amount of ice added to the water was close to

(Specific heat of water = 4.2 {Jg}^{- 1}^{\circ} C^{- 1} Specific heat of Ice = 2.1 {Jg}^{- 1}^{\circ} C^{- 1} Heat of fusion of water at 0^{\circ} C = 334 {Jg}^{- 1})

Q. The amount of heat required to raise the temperature of

75 g

of ice at

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to water at

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is:

[latent heat of fusion of ice is

80 c a l / g

, specific heat of water is

1 c a l / g^{o} C

]

Q. If 10 g of ice is added to 40 g of water at

15^{o} C

, then the temperature of the mixture is: (specific heat of water

= 4.2 \times 10^{3} J {k g}^{- 1} K^{- 1}

, Latent heat of fusion of ice

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

)

Q. X g of ice at

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340

g of water at

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C. The final temperature of the resultant mixture is

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= 333 J / g

; Specific heat of water

= 4.184 J / (g . K)

2

kg ice at

- 20^{o} C

is mixed with

5

kg water at

20^{o} C

. The final amount of water in the mixture would be ____.
Given the specific heat of ice =

0.5 c a l / g^{C}

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; Latent heat of fusion for ice =

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Ice at 20Cis added to 50g of water at 40C.When the temperature of the mixture reaches 0C,it is found that 20g of ice is still unmelted. The amount of ice added to the water was close to (specific heat of water =4.2J/g/oC)Specific heat of Ice =2.1J/g/oC Heat of fusion of water at 0oC=334j/g

The correct option is D 40gLet the amount of ice is m gm.According to the principal of calorimeterheat taken by ice = heat given by water∴20×2.1×m+(m−20)×334=50×4.2×40376m=8400+6680m=40.1

The correct option is D $40 g$
Let the amount of ice is m gm.
According to the principal of calorimeter
heat taken by ice = heat given by water
$∴ 20 \times 2.1 \times m + (m - 20) \times 334$
$= 50 \times 4.2 \times 40$
$376 m = 8400 + 6680$
$m = 40.1$