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Heating Curve of Ice

How much heat...

Question

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 $g^{- 1}$ .

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Solution

Heat released when 5 g water at 20°C turns to 5 g water at 0°C is,

H1 = (5)(4.2)((20 – 0)

=> H1 = 420 J

Heat released when 5 g of water at 0°C turns to 5 g of ice at 0°C.

H2 = (5)(336) = 1680 J

So, total heat released is = H1 + H2 = 2100 J

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Similar questions

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.

100 g

- 10^{o} C

100^{o} C

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

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

= 336 J g^{- 1}

10 g

- 10^{\circ} C

10 g

10^{\circ} C

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

336 J g^{- 1}

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

250 g

0^{\circ} C

0^{\circ} C

20^{\circ} C

336 J g^{- 1}

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

100 g

0^{\circ} C

0^{\circ} C

2

0^{\circ} C

20^{\circ} C

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

= 336 J g^{- 1}]

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