I jave a numerical . if anyone can answer that will be a great help for me.
350g of ice at 30c is contained in a copper vessel of mass 50g . Calculate the mass of the ice required to bring down the temperature of vessel and its content to 5c.Given: Specific heat capacity of water=4200 Specific heat capacity of copper =400 and specific latent heat of fusion of ice= 336000.

Open in App

Solution

Ice is kept at -30 C and it is heated (by the contained) till an equilibrium of 5 C is reached
so, Initial Temperature = -30 C
Final Temperature = 5 C

Here we can use the following relation
Heat Lost by Contained = Heat Gained by Ice
and
Heat Lost by container = Qc = m1c(dt)
Heat Gained by Ice as it converts to water = mc1(dt1) + mL + mc2*(dt2)

where m is the mass of the copper contained and m1 is the required mass of ice melted
so,
**m1c(dt) = mc1(dt1) + mL + mc2*(dt2)**

by substituting in the required values we can get the answer

Suggest Corrections

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

Q. A copper vessel of mass 100g contains 150g of water at

50^{o} C

. How much ice is needed to cool it to

5^{o} C

?
Given: Specific heat capacity of copper

= 0.4 J g^{- 1}

^{o} C^{- 1}

Specific heat capacity of water

= 4.2 J g^{- 1}

^{o} C^{- 1}

Specific latent heat of fusion of ice

= 336 J g^{- 1}

A vessel of negligible heat capacity contains 40 g of ice in it at 0⁰C. A metal of mass 50 g at 50⁰C is placed into the ice to melt it. The final temperature of the contents of the vessel is 10⁰C. Find the specific capacity of the metal.

(Specific latent heat of fusion of ice = 336 J/g and Specific heat capacity of water = 4.2 J/g⁰C)

Calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30°C, such that the final temperature of the mixture is 5^oC. Take specific heat capacity of material of vessel as 0.4 J g^-1 ^oC^-1 ; specific latent heat of fusion of ice = 336 J g^-1 ; specific heat capacity of water = 4.2 J g^-1 °C^-1.

Join BYJU'S Learning Program