1-6 kg of boiling water at 100 deg-c is poured into 2 kg of crushed ice at 0 deg-c- such that final temp- recorded is 0 deg-c- Calculate the specific latent heat of ice- - sp- latent heat of water - 4200 J-kg-K -

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Byju's Answer

Standard VII

Chemistry

Evaporation

1.6 kg of boi...

Question

1.6 kg of boiling water at 100°c is poured into 2 kg of crushed ice at 0°c, such that final temp. recorded is 0°c. Calculate the specific latent heat of ice? ( sp. latent heat of water = 4200 J/kg/K )

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Solution

Dear Student

Boiling water is at higher temperature and crushed ice is lower temperature.
So boiling water looses heat energy
$H e a t l o s t = h e a t g a i n e d m_{w} c_{w} ∆ t = m_{i c e} L (1.6) (4200) (373 K - 273 K) = (2) L L = \frac{(1.6) (4200) (100)}{2} = 336000 J / K G - K$

Regards

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Given: Specific heat capacity of water=4200 J k $g^{- 1} K^{- 1}$ ,specific latent heat of ice= $336 \times 10^{3} J k g^{- 1}$ .

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into steam at

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. Specific heat capacity of ice is

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, latent heat of fusion of ice

= 3.36 \times 10^{5} J {k g}^{- 1}

, specific heat capacity of water

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and latent heat of vapourization of water

= 2.25 \times 10^{6} J {k g}^{- 1}

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into water at

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? (specific latent of heat of fusion of water =

3, 34, 000 J / K g

, specific heat capacity of water =

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

A piece of ice of mass 40 g is dropped into 200 g of water at 50 $^{\circ} C$ . Calculate the final temperature of the water after all the ice has melted. (Specific heat capacity of water = 4200 J/kg $^{\circ} C$ , specific latent heat of fusion of ice $= 336 \times 10^{3}$ J/kg)

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, Latent heat of fusion of ice is

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)

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1.6 kg of boiling water at 100°c is poured into 2 kg of crushed ice at 0°c, such that final temp. recorded is 0°c. Calculate the specific latent heat of ice? ( sp. latent heat of water = 4200 J/kg/K )

Dear Student Boiling water is at higher temperature and crushed ice is lower temperature. So boiling water looses heat energy Heat lost = heat gainedmwcw∆t=miceL(1.6)(4200)(373K-273K)=(2)LL=(1.6)(4200)(100)2=336000 J/KG -K Regards

Dear Student

Boiling water is at higher temperature and crushed ice is lower temperature.
So boiling water looses heat energy
$H e a t l o s t = h e a t g a i n e d m_{w} c_{w} ∆ t = m_{i c e} L (1.6) (4200) (373 K - 273 K) = (2) L L = \frac{(1.6) (4200) (100)}{2} = 336000 J / K G - K$

Regards