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?

Question

Blaine Schmidt

Physics

13 April, 05:15

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?

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Answers (1)

Know the Answer?

Ryland Carlson · Answer 1

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.