1 kg of ice at 0 0 C is mixed with 1 kg of steam at 100 0 C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.36*10 5 J/kg and latent heat of vaporization of water=2.26*10 6 J/kg.

Question

Babulal Prasad · Answer

heat needed to melt the ice = mL = 1* 3.36*100000 =336000 J heat needed to increase the temperature of water to 100 degree centigrade=1*4200=4200 J total heat requirement= 340200 suppose m mass of steam is converted in to water then m*2260000= 340200 m= 340200/2260000=150.5 g final composition water=1150.5 g & steam=849.5 g

Vidya Iyer · Answer

Energy required to melt ice completely= 1000*80 cal Energy required to raise it's temperature to 100°= 1000*1*100 Total energy=1000(80+100) cal Mass of steam required= m m*540*1000 =180*1000 m= 0.33kg

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem below. Heat needed to melt the ice = mL = 1* 3.36*100000 =336000 J Heat needed to increase the temperature of water to 100 degree centigrade=1*4200=4200 J Total heat requirement= 340200 J Suppose m mass of steam is converted in to water then, m*2260000= 340200 m = 340200/2260000 = 150.5 g Final composition water = 1150.5 g & steam=849.5 g Thanks and regards, Kushagra

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1 kg of ice at 0 0 C is mixed with 1 kg of steam at 100 0 C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.3610 5 J/kg and latent heat of vaporization of water=2.2610 6 J/kg.

1 kg of ice at 0⁰C is mixed with 1 kg of steam at 100⁰C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.3610⁵ J/kg and latent heat of vaporization of water=2.2610⁶ J/kg.

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1 kg of ice at 0 0 C is mixed with 1 kg of steam at 100 0 C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.36*10 5 J/kg and latent heat of vaporization of water=2.26*10 6 J/kg.

1 kg of ice at 00C is mixed with 1 kg of steam at 1000C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.36*105 J/kg and latent heat of vaporization of water=2.26*106 J/kg.

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1 kg of ice at 0 0 C is mixed with 1 kg of steam at 100 0 C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.3610 5 J/kg and latent heat of vaporization of water=2.2610 6 J/kg.

1 kg of ice at 0⁰C is mixed with 1 kg of steam at 100⁰C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.3610⁵ J/kg and latent heat of vaporization of water=2.2610⁶ J/kg.