JEE Main202522 Jan 2025Morning ShiftPhysicsThermal Properties of MatterActual
An amount of ice of mass 10⁻³ ~kg and temperature -10^ C is transformed to vapour of temperature 110^ C 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 10^5 Jkg ⁻¹ and Latent heat of steam =2.25 10^6 Jkg ⁻¹ )
Options
- A3043 J
- B3024 J
- C3003 J
- D3022 J
Correct answer
A. 3043 J
Step-by-step solution
aligned & Q ₁= m S ₁ T =10⁻³ 2100 10=21 ~J & Q ₂= m L _ f =10⁻³ 3.35 10^5=335 ~J & Q ₃= m S _ w T =10⁻³ 4180 100=418 ~J & Q ₄= m L _ v =10⁻³ 2.25 10^6=2250 ~J & Q ₅= m S _ v T =10⁻³ 1920 10=19.2 ~J & Q _ net =3043.2 ~J aligned