Calculate the amount of heat required to convert 1-00 kg of ice at -10- C into steam at 100- C at normal pressure- S ice - -2100 J kg -1 K -1- S water -4200 J kg -1 K -1 Lf ice -3-36 - 105 J kg -1 Lv water -2-25 - 106 J kg -1A- 3-03 - 106 J- 4-03 - 106 J- 5-03 - 106 JD- 5-03 - 106 J

The correct option is A 3.03 × 10^6 JQ1: ice (at 10^∘ C) to ice (at 0^∘ C) = 1 kg × 2100 J kg^ 1 k^ 1× 10 = 21000 J Q2: ice (0^∘ C) to water (0^∘ C) = ...

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Standard X

Physics

Phase Change

Calculate the...

Question

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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5.03 \times 10^{6} J

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Solution

The correct option is A $3.03 \times 10^{6} J$
Q1: $i c e (a t - 10^{\circ} C) t o i c e (a t 0^{\circ} C) = 1 k g \times 2100 J k g^{- 1} k^{- 1} \times 10 = 21000 J$
Q2: $i c e (0^{\circ} C) t o w a t e r (0^{\circ} C) = 1 k g \times 3.36 \times 10^{5} J k g^{- 1} = 336000 J$
Q3: $w a t e r (0^{\circ} C) t o w a t e r (100^{\circ} C) = 1 k g \times 4200 J k g^{- 1} k^{- 1} \times 100 = 420000 J$
Q4: $w a t e r (100^{\circ} C) t o s t e a m (100^{\circ} C) = 1 k g \times 2.25 \times 10^{6} J k g^{- 1} = 2250000 J$

Net $Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$= 3.03 \times 10^{6} J$

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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})

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}

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

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 Lfice=3.36×105 J kg−1 Lvwater=2.25×106 J kg−1)