You cool a 130.0 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. how do you do this? part a what is the final temperature of the water? part b what is the final mass of the iron and the remaining water?

A) 100°C

B) 211 g

Explanation:

Heat released by red hot iron to cool to 100°C = 130 x .45 x 645 [ specific heat of iron is .45 J /g/K]

= 37732.5 J

heat required by water to heat up to 100 °C = 85 x 4.2 x 80 = 28560 J

As this heat is less than the heat supplied by iron so equilibrium temperature will be 100 ° C. Let m g of water is vaporized in the process . Heat required for vaporization = m x 540x4.2  = 2268m J

Heat required to warm the water of 85 g to 100 °C = 85X4.2 X 80 = 28560 J

heat lost = heat gained

37732.5 = 28560 + 2268m

m = 4 g.

So  4 g of water will be vaporized and remaining 81 g of water and 130 g of iron that is total of 211 g will be in the cup . final temp of water will be 100 °C.

it may be wrong but i think its a

0.55x745.7j and a calculator