The heat energy required to vaporize $5$ kg of water at $373$ K is nearly (Latent heat of vaporization $L_{v} = 2270 J$ )

2700 K.cal

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1000 K.cal

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27 K.cal

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270 K.cal

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Solution

The correct option is A 2700 K.cal
Temperature of water is 373K i.e. 100 $^{o}$ C
For the water to evaporate, heat energy required is equal to the latent heat of vaporization.
Latent heat of vaporization $= m \times L_{v} = 5 \times 2270 = 11350 k J = 2711.42 k C a l$

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Similar questions

Q. The heat energy required to vapourise 5kg of water at 373 K is

Q. The amount of heat required to convert

5 g

of ice at

0^{\circ} C

5 g

of steam at

100^{\circ} C

is -
[Latent heat of vaporization and fusion are

L_{v} = 540 {cal g}^{- 1}

and

L_{f} = 80 {cal g}^{- 1}

]

Ratio of heat energy required to change in states of $100 g$ of water at $100^{\circ} C$ to steam to $540 g$ of ice to water respectively is $5 : y$ then $y$ = ___

Take latent heat of fusion $80 c a l / g m$ and latent heat of vaporization as $540 c a l / g m$

Q. Calculate the entropy change, involved in the conversion of one mole of liquid water at

373 K

to vapour at same temperature is ( Latent heat of vaporization

= 2.275 K J / g

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

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The heat energy required to vaporize 5 kg of water at 373 K is nearly (Latent heat of vaporization Lv=2270 J)

The correct option is A 2700 K.calTemperature of water is 373K i.e. 100oCFor the water to evaporate, heat energy required is equal to the latent heat of vaporization.Latent heat of vaporization=m×Lv=5×2270=11350kJ=2711.42kCal

The heat energy required to vaporize $5$ kg of water at $373$ K is nearly (Latent heat of vaporization $L_{v} = 2270 J$ )

The correct option is A 2700 K.cal
Temperature of water is 373K i.e. 100 $^{o}$ C
For the water to evaporate, heat energy required is equal to the latent heat of vaporization.
Latent heat of vaporization $= m \times L_{v} = 5 \times 2270 = 11350 k J = 2711.42 k C a l$