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Thermodynamics Part 5

Section: Physics

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Question : 64 of 86

Marks: +1, -0

An amount of ice of mass

10^{-3} \mathrm{kg}

and temperature

-10^{\circ} \mathrm{C}

is transformed to vapour of temperature

110^{\circ} \mathrm{C}

by applying heat. The total amount of work required for this conversion is,

(Take, specific heat of ice

=2100 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}

, specific heat of water

4180 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}

, specific heat of steam

=

1920 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}

, Latent heat of ice

=3.35 \times 10^5 \mathrm{Jkg}^{-1}

and Latent heat of steam

=2.25 \times 10^6 \mathrm{Jkg}^{-1}

[22 Jan 2025 Shift 1]

3022 J
3043 J
3024 J
3003 J

Solution: 👈: Video Solution

ice at

-10^{\circ} \mathrm{C} \rightarrow

ice at

0^{\circ} \mathrm{C} \rightarrow

water at

0^{\circ} \mathrm{C}

→ vapour at

110^{\circ} \mathrm{C} \rightarrow

vapour at

100^{\circ} \mathrm{C} \rightarrow

water at

100^{\circ}

Q\;=10^{-3}\left[ 2100 \times 10 + 3.35 \times 10^5 + 100 \times 4180 + 2.25 \times 10^6 + 10 \times 1920 \right]

\;=3043.2 \mathrm{J}

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