You cool a 120.0 g slug of red-hot iron (temperature 745 ∘C) by dropping it into an insulated cup of negligible mass containing 85.0 g of water at 20.0 ∘C. Assume no heat exchange with the surroundings. What is the final temperature of the water?
What is the final mass of the iron and the remaining water?
Express your answer using three significant figures.

Explanation:

Heat lost by hot slug = gain gained by water

If we take that equilibrium temperature be boiling point is 100°C

Heat gained by water = 85 x 10⁻³ x 4.2 x 10³ x ( 100 - 20 )

= 28560 J

Heat lost by slug

= 120 x 10⁻³ x 460.55 x (745-100)

= 35646.570 J

So some water will be evaporated . Let it be m kg

35646.570 = 28560 + m x 2268 x 10³

m = 3.12 x 10⁻³ kg

= 3.12 g

3.12 g of water will be evaporated .

final mass = 120 + 85 - 3.12

= 201.88 g

= 202 g

0.11 m/s and 0.28 m/s should be correct