1 kg of ice at 0∘C is mixed with 1 kg of steam at 100∘C. Find the mass of steam when the mixture has reached thermal equilibrium. [Lfusion, ice=3.36×105 J/kg, Lvaporization, water=2.26×106 J/kg, Cwater=4200 J/kg-K]

1 kg of ice at 0∘C is mixed with 1 kg of steam at 100∘C. Find the

1.	1 kg of ice at 0∘C is mixed with 1 kg of steam at 100∘C. Find the mass of steam when the mixture has reached thermal equilibrium. [Lfusion, ice=3.36×105 J/kg, Lvaporization, water=2.26×106 J/kg, Cwater=4200 J/kg-K]
Answer» $1 kg$ of ice at $0^{\circ} C$ is mixed with $1 kg$ of steam at $100^{\circ} C$ . Find the mass of steam when the mixture has reached thermal equilibrium. $[L_{f u s i o n, i c e} = 3.36 \times 10^{5} J/kg,$ $L_{v a p o r i z a t i o n, w a t e r} = 2.26 \times 10^{6} J/kg,$ $C_{w a t e r} = 4200 J/kg-K]$

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1 kg of ice at 0∘C is mixed with 1 kg of steam at 100∘C. Find the mass of steam when the mixture has reached thermal equilibrium. [Lfusion, ice=3.36×105 J/kg, Lvaporization, water=2.26×106 J/kg, Cwater=4200 J/kg-K]

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