Calculate the amount of heat required to convert 1-00 kg of ice at -10- Cat normal pressure to steam at 100- C- The specific heat capacity of ice -2100 J kg -1 K -1 Latent heat of fusion of ice -3-36 - 105 J kg -1 The specific heat capacity of water -4200 J kg -1 K -1 Latent heat of vaporization of water -2-25 - 106 J kg -1A- 3-03 - 1012 JB- 3-303 - 106 J- 3-303 - 105 JD- 3-03 - 106 J

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Byju's Answer

Standard X

Physics

Latent Heat

Calculate the...

Question

$Calculate the amount of heat required to convert 1.00 kg of ice at -10°C$
$at normal pressure to steam at 100°C.$

$The specific heat capacity of ice = 2100 J k g^{- 1} K^{- 1}$

$Latent heat of fusion of ice = 3.36 \times 10^{5} J k g^{- 1}$

$The specific heat capacity of water = 4200 J k g^{- 1} K^{- 1}$

$Latent heat of vaporization of water = 2.25 \times 10^{6} J k g - 1$

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Solution

The correct option is B
$Heat required to take the ice from -10°C to 0°C$
$= (1 k g) (2100 J k g^{- 1} K^{- 1}) (10 K) = 21000 J .$
$The heat required to melt the ice at 0°C to water$
$= (1 k g) (3.36 \times 10^{5} J k g^{- 1}) = 336000 J .$
$Heat required to take 1 kg of water from 0°C to 100°C$
$= (1 k g) (4200 J k g^{- 1} K^{- 1}) (100 K) = 420000 J .$
$Heat required to convert 1 kg of water of 100°C into steam$
$= (1 k g) (2.25 \times 10^{6} J k g^{- 1}) = 2250000 J$

So, total heat required $= 3.03 \times 10^{6} J$

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Similar questions

Q. Calculate the amount of heat required to convert 1.00 kg of ice at

- 10^{\circ}

C into steam at

100^{\circ}

C at normal pressure.

(S_{i c e} = 2100 J k g^{- 1} K^{- 1},

S_{w a t e r} = 4200 J k g^{- 1} K^{- 1},

L_{f, i c e} = 3.36 \times 10^{5} J k g^{- 1}

L_{v, w a t e r} = 2.25 \times 10^{6} J k g^{- 1})

The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 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 $g^{- 1} K^{- 1}$ .

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

Q. Calculate the heat required to convert 

3 k g

of ice

a t - 12^{\circ} C

kept in a calorimeter to steam at

100^{\circ}

at atmospheric pressure. Given : specific heat of ice

= 2100 \times 10^{5} J {k g}^{- 1} K^{- 1}

, specific heat of water

= 4186 \times 10^{3} J {k g}^{- 1} K^{- 1}

, latent heat of fusion of

c e = 3.35 \times 10^{5} J {k g}^{- 1}

and latent heat of steam

= 2.256 \times 10^{6} J {k g}^{- 1}

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$

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