20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The minimum value of m so that finally all ice and steam converts into water at 0 ∘C is (Use: specific heat and latent heat as Cice=0.5 cal/gm ∘C,Cwater=1 cal/gm ∘C,Lfusion =80 cal/gm and Lvapor =540 cal/gm )

20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The mini

1.	20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The minimum value of m so that finally all ice and steam converts into water at 0 ∘C is (Use: specific heat and latent heat as Cice=0.5 cal/gm ∘C,Cwater=1 cal/gm ∘C,Lfusion =80 cal/gm and Lvapor =540 cal/gm )
Answer» $20 gm$ ice at $- 10^{\circ} C$ is mixed with $m gm$ steam at $100^{\circ} C$ . The minimum value of $m$ so that finally all ice and steam converts into water at $0^{\circ} C$ is $($ Use: specific heat and latent heat as $C_{i c e} = 0.5 cal/gm^{\circ} C, C_{water} = 1 cal/gm^{\circ} C, L_{f u s i o n}$ $= 80 cal/gm$ and $L_{v a p o r}$ $= 540 cal/gm$ $)$

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20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The minimum value of m so that finally all ice and steam converts into water at 0 ∘C is (Use: specific heat and latent heat as Cice=0.5 cal/gm ∘C,Cwater=1 cal/gm ∘C,Lfusion =80 cal/gm and Lvapor =540 cal/gm )

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