250 g of water at 30-C is contained in a copper vessel of mass 50 g- Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5-C- Given - specific latent heat of fusion of ice - 336 - 103 J kg-1- specific heat capacity of copper - 400 J kg-1 K-1- specific heat capacity of water - 4200 J kg-1 K-1 -

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Standard XII

Physics

Specific Heat Capacity

250 g of wate...

Question

$250 g$ of water at $30^{\circ} C$ is contained in a copper vessel of mass $50 g$ . Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $5^{\circ} C$ .
Given : specific latent heat of fusion of ice = $336 \times 10^{3} J k g^{- 1}$ , specific heat capacity of copper = $400 J k g^{- 1} K^{- 1}$ , specific heat capacity of water = $4200 J k g^{- 1} K^{- 1}$ :

7961 g

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7.961 g

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74.93 g

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85.61 g

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Solution

The correct option is C $74.93 g$
Heat required to bring down the temperature to $5^{o}$ is
$250 \times 4.2 \times (5 - 30) + 50 \times 0.4 \times (- 25)$
$= - 26750 J$

Let $m$ be the mass of ice so,
Heat released by ice is
$m \times 336 + m \times 4.2 \times 5$
$357 m J$
Now the energy of the system must remain conserved so,
$357 m - 26750 = 0$
$∴ m = 74.93 g$

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Similar questions

250 g of water at 30 °C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 °C.

Specific latent heat of fusion of ice = 336 x 10³ J kg^-1

Specific heat capacity of copper vessel= 400 J kg^-1 °C^-1

Specific heat capacity of water 4200 J kg^-1 °C^-1

Q 8)

250 g of water at 30 degree celsius is contained in a copper vessel of mass 50g.Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 degeree celsius.Specific latent heat of fusion of ice 336*1000 J per kg.

Specific heat capacity of copper =400 J per kg per K Specific heat Capacity of water =4200J per kg per K.

Q. A piece of ice of mass

40 g

0^{o} C

is added to

200 g

of water at

50^{\circ} C

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

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

and specific latent heat of fusion of ice =

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

40 g

of ice at

0^{o} C

is used to bring down the temperature of certain mass of water at

60^{o} C t o 10^{o} C

. Find the mass of water used.
[Specific heat capacity of water

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

[Specific latent heat of fusion of ice

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

]

The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 7,77,000 J. Calculate the specific latent heat of ice.

Specific heat capacity of ice = 2100 J $k g^{- 1} K^{- 1}$ , specific heat capacity of water=4200 J k $g^{- 1} K^{- 1}$ .

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The correct option is C 74.93 gHeat required to bring down the temperature to 5o is250×4.2×(5−30) +50×0.4×(−25)=−26750JLet m be the mass of ice so, Heat released by ice ism×336 +m×4.2×5357m JNow the energy of the system must remain conserved so,357m−26750=0∴m=74.93 g