ΔQ1=m×S1×ΔT=10−3×2100×10=21 JΔQ2=m×Lf=10−3×3.35×105=335 JΔQ3=m×Sw×ΔT=10−3×4180×100=418 JΔQ4=m×Lv=10−3×2.25×106=2250 JΔQ5=m×Sv×ΔT=10−3×1920×10=19.2 JΔQnet=3043.2 J
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 =2100Jkg−1 K−1, specific heat of water =4180Jkg−1 K−1, specific heat of steam =1920Jkg−1 K−1, Latent heat of ice =3.35×105Jkg−1 and Latent heat of steam =2.25×106 Jkg−1 )
Held on 22 Jan 2025 · Verified 6 Jul 2026.
3043 J
3024 J
3003 J
3022 J
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