A 7.7-kg cube of copper (c Cu = 386 J/kg-K) has a temperature of 750 K. It is dr

Question

A 7.7-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.4 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?

K

Your submissions:

338.65

Computed value:

338.65

Submitted:

Monday, April 25 at 9:52 PM

Feedback:

2)

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?

kg

Your submissions:

1.226

Computed value:

1.226

Submitted:

Monday, April 25 at 9:54 PM

Feedback:

3)

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 ?

Explanation / Answer

3)
Specific Heat of Copper = 0.386 J/gm k
Mass of copper = 7.7 kg = 7.7 * 10^3 gm
Amount of Heat Energy needed by copper to lower temperature by, 750 K - 373 K = 377 K
Latent heat of Vaporisation , = 2257 J/gm
Q = m*Cc*T
Q = 7.7 * 10^3 * 0.386 * 377 J
Q = 1.12 * 10^6 J

Let the mass of water to be added = mw grams

Heat given by water, Q = mw*cw*T + mw * 2257

Heat Lost by water = Heat gained by Copper
mw*cw*T + mw * 2257 = 1.12 * 10^6 J
mw * 4.186 * (373 - 293) + mw * 2257 = 1.12 * 10^6 J
mw = 432.1 gm

Mass of water to be added, mw = 0.432 Kg

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