if 20g of steam at 100 c is mixed into 80 g of water at 20 c what is the final temperature if no heat lost

Answer 1

I answered this before but the question disappeared

Heat out of steam= 20 {2260+4.2(100-T)}

Heat into water = 80 (T-20)

so

4(T-20) = 2260 + 4.2(100-T)

4 T - 80 = 2260 + 4200 - 4.2 T

8.2 T = 6540

T = 797 which is silly of course
What it means is that 80 grams of water at 20 degrees simply can not condense 20 grams of steam. The final temp is 100 deg C and you still have some stem left :)