In a container of negligible mass 30g of steam at `100^@C` is added to 200g of water that has a temperature of `40^@C` If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take `L_v = 539 cal//g and c_(water) = 1 cal//g-^@C.`

In a container of negligible mass 30g of steam at `100^@C` is added to

1.	In a container of negligible mass 30g of steam at `100^@C` is added to 200g of water that has a temperature of `40^@C` If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take `L_v = 539 cal//g and c_(water) = 1 cal//g-^@C.`
Answer» Correct Answer - A::B::C::D Let Q be the heat required to convert 200 g of water at `40^@C "into" 100^@C`, then `Q= mcDeltaT = (200)(1.0)(100-40) = 12000 cal` Now, suppose `m_0` mass of steam converts into water to liberate this much amount of heat, then `m_0 = Q/L = 12000/539 = 22.26 g ` Since, it is less than 30g, the temperature of the mixture is `100^@C`. Mass of steam in the mixture `= 30- 22.26 = 7.74g ` and mass of water in the mixture `= 200 + 22.26 = 222.26g `

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