How much heat energy is released when 5-0 g water at 42-C changes into ice at 0-C- Take specific heat capacity of water - 4-2 Jg-1- specific latent heat of fusion of ice - 336 Jg-1 -

The correct option is D 2562 J Heat released Q=mL + mc(42 0) Q=5× 336+5× 4.2× 42 Q=2562 J ...

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How much heat energy is released when $5.0 g$ water at $42^{\circ} C$ changes into ice at $0^{\circ} C$ ? Take specific heat capacity of water = $4.2 J g^{- 1}$ , specific latent heat of fusion of ice = $336 J g^{- 1}$ :

21 J

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4200 J

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42 J

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2562 J

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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.

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}]

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}

2 K g

0^{0} C

20^{0} C

3, 34, 000 J / K g

4200 J {K g}^{- 1} K^{- 1}

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