350 g of water at 30 degree 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 degree C. given specific latent heat of fusion of ice=336000 J K^-1, specific heat capacity of copper=400 J K^-1, specific heat capacity of water=4200 J K^-1.

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Solution

Suppose the mass of ice required to bring down the temperature of the vessel and its contents is x.
Then ,

Heat lost =Heat gained
M x C1 x (30-5)^oC + M' x C2x (30-5)^oC = x L + x C1(5-0)^oC
M=mass of water,
M'=mass of vessel

Now putting the values in the terms of unit cal/g^oC, you can get the answer.

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Similar questions

250 g

of water at

30^{o} 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^{o} 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}

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

10

g of ice at 0

^{o}

C absorbs

5460

J of heat energy to melt and change to water at

50

^{o}

C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is

4200

J kg

^{- 1}

^{- 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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