Thermodynamics: JEE Main 2025 Physics question

JEE Main 2025 · 22 Jan, Morning shift · Q27

JEE Main 202522 Jan, Morning shiftPhysicsMediumThermodynamics163× since 2002

An amount of ice of mass

10^{-3}

kg and temperature

-10^\circ

C is transformed to vapour of temperature

110^\circ

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

=2100

J kg

^{-1}

^{-1}

, specific heat of water

=4180

J kg

^{-1}

^{-1}

, specific heat of steam

=1920

J kg

^{-1}

^{-1}

, Latent heat of ice

=3.35\times10^{5}

J kg

^{-1}

and Latent heat of steam

=2.25\times10^{6}

J kg

^{-1}

)

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JEE Main 2025 Thermodynamics Question with Solution (22 Jan Morning shift, Q27) | JEE Arc

JEE Main 2025 · 22 Jan, Morning shift · Q27

JEE Main 202522 Jan, Morning shiftPhysicsMediumThermodynamics163× since 2002

An amount of ice of mass

10^{-3}

kg and temperature

-10^\circ

C is transformed to vapour of temperature

110^\circ

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

=2100

J kg

^{-1}

^{-1}

, specific heat of water

=4180

J kg

^{-1}

^{-1}

, specific heat of steam

=1920

J kg

^{-1}

^{-1}

, Latent heat of ice

=3.35\times10^{5}

J kg

^{-1}

and Latent heat of steam

=2.25\times10^{6}

J kg

^{-1}

)