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-1 K-1- Swater - 4200 J kg-1 K-1 Lfice-3-36 - 105 J kg-1 Lvwater - 2-25 - 106 J kg-1

Q1: 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) = 1 kg × 3.36 × 10^5 J kg^ 1 = 336000 J ...

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Standard XI

Physics

Calorimetry

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

1 kg

of ice at

0^{\circ} C

is mixed with

1 kg

of steam at

100^{\circ} C

. Find the mass of steam when the mixture has reached thermal equilibrium.

[L_{f u s i o n, i c e} = 3.36 \times 10^{5} J/kg,

L_{v a p o r i z a t i o n, w a t e r} = 2.26 \times 10^{6} J/kg,

C_{w a t e r} = 4200 J/kg-K]

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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)