A 8.8-kg cube of copper (c_Cu = 386 J/kg-K) has a temperature of 750 K. It is dr

Question

A 8.8-kg cube of copper (c_Cu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.5 kg of water (c_water = 4186 J/kg-K) with an initial temperature of 293 K. What is the finat temperature of the water-and-cube system? 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 times 10^6 J/kg) will be left after the water stops boiling? kg Let's try this again, but this time add just the minimum amount of water needed to tower 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? kg

Explanation / Answer

Cw = 4186 J/kg-K , Cc = 386 J/kg-K

mass of copper mcu = 8.8 kg

initial temperature of copper Tc = 750 K

mass of water mw = 5.5 kg

initial temperature of water Tw = 293 K

1) Heat gained by water = Heat lost by copper

mw Cw (T-Tw ) = mc Cc (Tc-T)

5.5 x 4186 x (T - 293) = 8.8 x 386 x (750 -T)

23023 T - 6745739 = 2547600 - 3396.8 T

26419.8 T = 9293339

Final temperature is T= 351.7 K

2)

initial temperature of copper Tc = 1350 K

Latenet heat of vaporization of water is Lw = 2.26 x 106 J/kg

m is mass water that gets evaporated.

final temperature attained T = 1000C = 373 K

Heat gained by water = Heat lost by copper

mw Cw (T-Tw ) + m L = mc Cc (Tc-T)

5.5 x 4186 x (373 - 293) + m x 2.26 x 106= 8.8 x 386 x (1350 - 373)

1841840 +  m x 2.26 x 106 = 3318673.6

m = 0.653 kg

mass of evaporated water m = 0.653 kg

mass of liquid water left after water stops boiling is mw - m = 5.5 - 0.653 = 4.85 kg

3)

initial temperature of copper Tc = 750 K

initial temperature of water Tw = 293 K

mass of copper mcu = 8.8 kg

mass of water is m to be calculated.

final temperature of the system T = 373 K

Heat gained by water = Heat lost by copper

m Cw (T-Tw ) + m L = mc Cc (Tc-T)

m x 4186 x (373 - 293) + m x 2.26 x 106= 8.8 x 386 x (750 - 373)

m x 2594880 =1280593.6

m = 0.493 kg

Amount of water we need is = 0.493 kg

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