Question

If 10 gm of ice at 0 is mixed with 50 gm of water at 12.What is the final temperature?

Answer 1

We have ice at -10°C,For the ice to come to 0°C, the energy required is m*s*∆Tm is mass of ice = 10gs is specific heat of ice = 0.8 cal/g.°C∆T is temperature difference = 0-(-10) = 10°CSo the energy required is 10*0.8*10 = 80 calThe energy released when 50 grams of water at 15°C becomes water at 0°C is m*s*∆Ts for water is 1 cal/g.°CEnergy released is 50*1*(15–0) = 50*15 = 750 calFor the ice to completely turn into water at 0°C, the energy required is m*LL is latent heat of ice = 80 cal/gEnergy required is 10*80 = 800 cal.But the water when turns into water at 0°C, the energy released is only 750 cal.So, the total ice will not get converted into water.The energy left after the ice gets to 0°C is 750–80=680 calWith 680 cal, the amount of ice that turns into water is 680/80 = 8.5 grams.So the composition is 10–8.5=1.5 grams of ice at 0°C and 10+8.5=18.5 grams of water at 0°C.