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

Byju's Answer

Standard X

Physics

Transformers

Calculate the...

Question

Calculate the amount of heat released when 5.0 g of water at 20 °C is changed into ice at 0 °C.

Specific heat capacity of water = 4.2J g^-1 °C^-1, Specific latent heat of fusion of ice = 336 J g^-1.

2100 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4200 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1680 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

420 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
2100 J

Heat released by water to change its temperature from at 20^oC to 0^oC

= m x c x ∆t

= [5 x 4.2 x (20-0)] J = 420 J.

Heat released when water at 0^oC is changed to ice at 0°C = m x L = (5 x 336) J = 1680 J.

Total heat released = (420 + 1680) = 2100 J

Hence, 2100 J of heat energy is released.

Suggest Corrections

Similar questions

How much heat energy is released when 5.0 g of water at $20^{o} C$ changes into ice at $0^{o} C$ ? Take specific heat capacity of water = 4.2 J $g^{-} 1 K^{- 1}$ , specific latent heat of fusion of ice = 336 J $g^{- 1}$ .

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

100 g

of ice at

- 10^{o} C

completely into water at

100^{o} 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}

250 g

of ice at

0^{\circ} C

converts into water at

0^{\circ} C

in a minute when a constant heat is supplied. In how much time (sec) the water thus obtained will change to

20^{\circ} C

? The specific latent heat of ice is

336 J g^{- 1}

. The specific heat capacity of water is

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

10 g of ice at 0 $^{\circ} C$ absorbs 5460 J of heat energy to melt and change t water at 50 $^{\circ} C$ . Calculate the specific latent heat of fusion of ice. (Specific heat capacity of water is 4200 J $K g^{- 1} K^{- 1}$ .

A refrigerator converts 100 g of water at $20^{o} C$ to ice at $- 10^{o} C$ in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J $g^{- 1} K^{- 1}$ , specific latent heat of ice is 336 J $g^{-} 1$ and the specific heat capacity of ice is 2.1 J $g^{- 1} K^{- 1}$ .

Join BYJU'S Learning Program

Calculate the amount of heat released when 5.0 g of water at 20 °C is changed into ice at 0 °C. Specific heat capacity of water = 4.2J g-1 °C-1, Specific latent heat of fusion of ice = 336 J g-1.

Calculate the amount of heat released when 5.0 g of water at 20 °C is changed into ice at 0 °C.

Specific heat capacity of water = 4.2J g^-1 °C^-1, Specific latent heat of fusion of ice = 336 J g^-1.