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Standard X

Physics

Principle of Calorimetry

Calculate the...

Question

Calculate the heat energy required to convert completely $10 k g$ of water at $50 ° C$ into steam at $100 ° C$ , given that the specific heat capacity of water is $4200 J / (k g ° C)$ and the specific latent heat of vaporisation of water is $2260 k J / k g$ .

30451 KJ

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60562 KJ

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72103 KJ

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24700 KJ

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Solution

The correct option is D 24700 KJ
Given that,

Mass $m = 10 k g$

We know that,

$Q_{1} = m c Δ t$

$Q_{1} = 10 \times 4200 \times (100 - 50)$

$Q_{1} = 2100 K J$

Now, the latent heat

$Q_{2} = m L$

$Q_{2} = 10 \times 2260$

$Q_{2} = 22600 k J$

Now, the total heat energy required

$Q = Q_{1} + Q_{2}$

$Q = 2100 + 22600$

$Q = 24700 K J$

Hence, this is the required solution

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Similar questions

Q. Calculate the increase in the internal energy of

20 g

of water when it is heated from

0^{\circ} C

100^{\circ} C

and converted into steam at

100 kPa

. The density of steam

= 0.50 {kg/m}^{3}

, specific heat capacity of water

= 4200 {J kg}^{- 1}^{\circ} C^{- 1}

and latent heat of vaporization of water

= 2.3 \times 10^{6} J/kg

Q. How much energy is required to change

2 K g

of ice at

0^{0} C

into water at

20^{0} C

? (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}

Q. The amount of heat energy required to convert

1

kg of ice at

- 10^{\circ} C

to water at

100^{\circ} C

7, 77, 000

J. Calculate the specific latent heat of ice. Specific heat capacity of ice

2100 J k g^{- 1} K^{- 1}

, specific heat capacity of water =

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

Q. The increase in the internal energy of

10 g

of water when it is heated from

0^{\circ} t o 100^{\circ}

and converted into steam at

100 k P a

is given as

x \times 10^{2} J

. Find

x

.
The density of steam

= 0.6 k g / m^{3} .

Specific heat capacity of water

= 4200 J / k g -^{\circ} C

and the latent heat of vaporization of water

= 2.5 \times 10^{6} J / k 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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Calculate the heat energy required to convert completely 10 kg of water at 50°C into steam at 100°C, given that the specific heat capacity of water is 4200 J/(kg°C) and the specific latent heat of vaporisation of water is 2260 kJ/kg.

The correct option is D 24700 KJGiven that, Mass m=10kg We know that, Q1=mcΔt Q1=10×4200×(100−50) Q1=2100KJ Now, the latent heat Q2=mL Q2=10×2260 Q2=22600kJ Now, the total heat energy required Q=Q1+Q2 Q=2100+22600 Q=24700KJ Hence, this is the required solution

Calculate the heat energy required to convert completely $10 k g$ of water at $50 ° C$ into steam at $100 ° C$ , given that the specific heat capacity of water is $4200 J / (k g ° C)$ and the specific latent heat of vaporisation of water is $2260 k J / k g$ .