An amount of ice of mass $10^{-3} \mathrm{~kg}$ and temperature $-10^{\circ} \mathrm{C}$ is transformed to vapour of temperature $110^{\circ}$ 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 $=2.25 \times 10^{6} \mathrm{Jkg}^{-1}$)

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Heat & Thermodynamics

An amount of ice of mass

1 0^{- 3} kg

and temperature

- 1 0^{\circ} C

is transformed to vapour of temperature

11 0^{\circ}

by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice

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

, specific heat of water

= 4180 Jkg^{- 1} K^{- 1}

, specific heat of steam

= 1920 Jkg^{- 1} K^{- 1}

, Latent heat of ice

= 3.35 \times 1 0^{5} Jkg^{- 1}

and Latent heat of steam

= 2.25 \times 1 0^{6} Jkg^{- 1}

)

Answer: (B)

Heating the ice from

- 10° C

to vapour at

110° C

happens in five stages.

Warming ice from

- 10° C

0° C

Q_{1} = m S_{i} Δ T = 1 0^{- 3} \times 2100 \times 10 = 21 J .

Melting the ice at

0° C

Q_{2} = m L_{f} = 1 0^{- 3} \times 3.35 \times 1 0^{5} = 335 J .

Warming water from

0° C

100° C

Q_{3} = m S_{w} Δ T = 1 0^{- 3} \times 4180 \times 100 = 418 J .

Vaporising the water at

100° C

Q_{4} = m L_{v} = 1 0^{- 3} \times 2.25 \times 1 0^{6} = 2250 J .

Heating the steam from

100° C

110° C

Q_{5} = m S_{v} Δ T = 1 0^{- 3} \times 1920 \times 10 = 19.2 J .

Total heat required:

Q_{n e t} = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 J \approx 3043 J .

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option