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?

Explanation:

a.

-The specific heat of iron is 0.45 J /g/K

-Heat released by red hot iron to cool to 100°C

#The heat required by water to heat up to 100°C :

#You notice that this heat is less that heat supplied by the iron slug so the equilibrium temperature will be 100 ° C.

#Let m be the grams of water vaporized during the prices:

#Heat required to warm the water of 85 g to 100 °C is calculated as:

#For an equilibrium:

Heat Lost= Heat Gained

Hence, the amount of water vaporized is 4 grams

b. Final mass of iron and remaining water is calculated as;

Hence, the final mass of the iron and the remaining water is 211 grams.

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