A 8.5-kg cube of copper (c Cu = 386 J/kg-K) has a temperature of 750 K. It is dr
ID: 1838775 • Letter: A
Question
A 8.5-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?
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?
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
Mass of cube of copper m = 8.5 kg
Specific heat of copper c = 386 J/kg-K
Initial temperature of copper t = 750 K
Mass of water M = 5.7 kg
Specific heat of cwater C =4186 J/kg-K
Initial temperature of water t ' = 293 K.
1) Let the final temperature of the water-and-cube system be T .
Heat lost by copper = heat gain by water
mc (750 - T ) = MC ( T - 293)
8.5 x386 x(750-T) = 5.7 x4186 x( T - 293 )
3281 (750-T ) = 23860.2 ( T - 293 )
750 - T = 7.2722 ( T - 293)
= 7.2722 T - 2130.7
8.2722 T = 2880.7
T = 348.24 K
(2).heat required to evoporate the water Q = MC(373 - 293 ) + ML
Where L = Latent heat of vaporisation of water = 22.6 x 10 5 J/kg
Since the temprature of the water vapour is = 373 K
Substitute values you get , Q = (5.7 x4186 x80 ) + ( 5.7 x 22.6 x 10 5 )
= 1.908 x10 6 + 12.882 x10 6
= 14.79 x10 6 J
Heat lost by copper when it reaches the 393 K is mc ( t" - 373)
= 8.5 x 386x(1350 -373 )
Q ' = 3.205x10 6 J
Q ' < Q .
So, the final temprature of the water cube combination is 373 K
remaining heat after all the water reaches the temprature of 373 K is Q " = Q ' - (1.908 x10 6 )
= ( 3.205x10 6 )-(1.908 x10 6 )
Q " = 1.297 x10 6 J
Amount of water evaporate for this Q " amount of heat is m " = Q " / L
m " = (1.297 x10 6 )/( 22.6 x 10 5 J/kg)
= 0.5738 kg
Therefore required answer is = M - m "
= 5.7 kg - 0.5738 kg
= 5.126 kg
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