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Standard XII

Chemistry

Molar and Specific Heat Capacity

How much heat...

Question

How much heat energy is released when $5 g$ of water at $20^{o} C$ changes to ice at $0^{o} C$ ?
[Specific heat capacity of water $= 4.2 J g^{- 1} C^{- 1}$ .
Specific latent heat of fusion of ice $= 336 J g^{- 1}$ ]

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Solution

Heat energy $= m c Δ T + m L$
where, $m$ = mass of water
$c$ = specific heat capacity of water
$L$ = specific latent heat of fusion of ice
$Δ T$ = change in temperature
Therefore, heat energy released $= 5 \times 4.2 \times 20 + 5 \times 336 = 420 + 1680$ $= 2100 J$

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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.

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}

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}

Q. A refrigerator converts

100 g

of water at

20^{o} C

to ice at -

10^{o} C

73.5 m i n

. 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}

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

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How much heat energy is released when 5g of water at 20oC changes to ice at 0oC?[Specific heat capacity of water =4.2Jg−1C−1. Specific latent heat of fusion of ice =336Jg−1]

Heat energy =mcΔT+mLwhere, m = mass of waterc = specific heat capacity of waterL = specific latent heat of fusion of iceΔT = change in temperatureTherefore, heat energy released =5×4.2×20+5×336=420+1680 =2100J

How much heat energy is released when $5 g$ of water at $20^{o} C$ changes to ice at $0^{o} C$ ?
[Specific heat capacity of water $= 4.2 J g^{- 1} C^{- 1}$ .
Specific latent heat of fusion of ice $= 336 J g^{- 1}$ ]

Heat energy $= m c Δ T + m L$
where, $m$ = mass of water
$c$ = specific heat capacity of water
$L$ = specific latent heat of fusion of ice
$Δ T$ = change in temperature
Therefore, heat energy released $= 5 \times 4.2 \times 20 + 5 \times 336 = 420 + 1680$ $= 2100 J$