50gof ice at -20-Cis heated on an electric heater- till it forms water at 100-C- Calculate -a Heat energy required to raise the temperature of ice to 0-Cb Heat energy required to melt the ice so as to form water at 0-C-c Heat energy required to bring water to the temperature of 100-C-Take sp- latent heat of fusion of ice 340Jg-1- heat capacity of water 4-2Jg-1-C-1 and sp- heat capacity of ice 2-1Jg-1-C-1-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Thermal Expansion

50gof ice at ...

Question

$50 g$ of ice at $- 20^{\circ} C$ is heated on an electric heater, till it forms water at $100^{\circ} C$ . Calculate :
(a) Heat energy required to raise the temperature of ice to $0^{\circ} C$
(b) Heat energy required to melt the ice so as to form water at $0^{\circ} C$ .
(c) Heat energy required to bring water to the temperature of $100^{\circ} C$ .
Take sp. latent heat of fusion of ice $340 J g^{- 1}$ . heat capacity of water $4 \cdot 2 J {g^{- 1}}^{\circ} C^{- 1}$ and sp. heat capacity of ice $2.1 J {g^{- 1}}^{\circ} C^{- 1}$ .

Open in App

Solution

Step 1: Given data
The quantity of ice is $50 g$ and take specific latent heat of fusion of ice $340 J g^{- 1}$ ,
Heat capacity of water $4.2 J {g^{- 1}}^{\circ} C^{- 1}$
Specific heat capacity of ice $2.1 J {g^{- 1}}^{\circ} C^{- 1}$ .
Step 2: Calculating (a) heat energy required to raise the temperature of ice to $0^{\circ} C$
Heat energy required to bring ice at $0^{\circ} C$
$= m c θ_{R} = 50 \times 2.1 \times (0 - (- 20)) = 50 \times 2.1 \times 20 = 2100 J$
Step 3: Calculating the (b) heat energy required to melt the ice so as to form water at $0^{\circ} C$
Heat energy required to melt ice
$H = m L = 50 g \times 340 {Jg}^{- 1} = 17000 J$
Step 4: Calculating (c) the heat energy required to bring water to the temperature of $100^{\circ} C$
Heat energy required to bring water to $100^{\circ} C$
$= m c θ_{R} = 50 g \times 4.2 {Jg}^{- 10} C^{- 1} \times 100 ° C = 21000 J .$
Therefore,
(a) The heat energy required to raise the temperature of ice to $0^{\circ} C$ is $2100 J .$
(b) The heat energy required to melt the ice so as to form water at $0^{\circ} C$ is $17000 J .$
(c) The heat energy required to bring water to the temperature of $100^{\circ} C$ is $21000 J .$

Suggest Corrections

Similar questions

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 $k g^{-} 1 K^{- 1}$ .

Q. Calculate the total amount of heat energy required to convert

100 g

of ice at

- 10^{\circ} C

completely into water at

100^{\circ} C

. Specific heat capacity of ice =

2.1 J g^{- 1} K^{- 1}

, specific heat capacity of water =

4.2 J g^{- 1} K^{- 1}

, specific latent heat of ice =

336 J g^{- 1}

Q. Heat energy is supplied at a constant rate to

100 g

of ice at

0^{\circ} C

. The ice is converted into water at

0^{\circ} C

2

minutes. How much time will be required to raised the temperature of water from

0^{\circ} C

20^{\circ} C

? [Given: specific heat capacity of water

= 4.2 J g^{- 1}^{\circ} C^{- 1}

, specific latent heat of ice

= 336 J g^{- 1}]

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}

Q. Calculate the mass of ice required to lower the temperature of

300 g

of water at

40 ° C

to water at

0 ° C

. (Specific latent heat of ice = 336 J /g, Specific heat capacity of water = 4.2 J/ g°C).

Join BYJU'S Learning Program