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. $(S_{i c e} = 2100 J k g^{- 1} K^{- 1},$
$S_{w a t e r} = 4200 J k g^{- 1} K^{- 1},$
$L_{f, i c e} = 3.36 \times 10^{5} J k g^{- 1}$ ,
$L_{v, w a t e r} = 2.25 \times 10^{6} J k g^{- 1})$

$3.03 \times 10^{6} J$

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$4.03 \times 10^{6} J$

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$5.03 \times 10^{6} J$

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$6.03 \times 10^{6} J$

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Solution

The correct option is A
$3.03 \times 10^{6} J$

Heat required to heat ice from $- 10^{\circ} C$ to ice at $0^{\circ} C$ is:
$Q_{1} = m S_{i c e} Δ T_{1}$
$Q_{1} = 1 k g \times 2100 J k g^{- 1} K^{- 1} \times 10 K$
$Q_{1} = 21000 J$

Heat required to covert ice to water is:
$Q_{2} = m L_{f, i c e}$
$Q_{2} = 1 k g \times 3.36 \times 10^{5} J k g^{- 1}$
$Q_{2} = 336000 J$

Heat required to heat water from $0^{\circ} C$ to water at $100^{\circ} C$ is:
$Q_{3} = m S_{w a t e r} Δ T_{3}$
$Q_{3} = 1 k g \times 4200 J k g^{- 1} k^{- 1} \times 100$
$Q_{3} = 420000 J$

Heat required to covert water to steam is:
$Q_{4} = m L_{v, w a t e r}$
$Q_{4} = 1 k g \times 2.25 \times 10^{6} J k g^{- 1}$
$Q_{4} = 2250000 J$

Total heat required is:
$Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$Q = (21 + 336 + 420 + 2250) k J$
$Q = 3.027 \times 10^{6} J$

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Similar questions

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

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

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

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

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

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

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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Calculate the amount of heat required to convert 1.00 kg of ice at −10∘C into steam at 100∘C at normal pressure. (Sice=2100 J kg−1K−1, Swater=4200 J kg−1K−1, Lf,ice=3.36×105 J kg−1, Lv,water=2.25×106 J kg−1)