A 8.6-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dro

Question

A 8.6-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.7 kg of water (cwater = 4186 J/kg-K) with an initial temperature of 293 K.

1) What is the final temperature of the water-and-cube system? (answer in K)

If the temperature of the copper was instead 1350 K, it would cause the water to boil. How much liquid water (latent heat of vaporization = 2.26 × 106 J/kg) will be left after the water stops boiling? (answer in kg)

Let's try this again, but this time add just the minimum amount of water needed to lower the temperature of the copper to 373 K. In other words, we start with the cube of copper at 750 K and we only add enough water at 293 K so that it completely evaporates by the time the copper reaches 373 K. Assume the resulting water vapor remaining at 373 K. How much water do we need in kg?

Explanation / Answer

1. Assuming no heat loss:

heat absorbed by water = heat released by copper

5.7 x 4.186 x (T - 293) = 8.6 x 0.386 x (750 - T)

23.86 T - 6991.04 = 2489.7 - 3.32 T

T = 348.8 K

2. heat released by copper cube

= 8.6 x 0.386 x (750 - 373) = 1251.49 kJ

heat absorbed by water

1251.49 = (m x 4.186 x 100) + (m x 2.26 x 10^3 )

1251.49 = 2678.6 m

m = 0.467 kg

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