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Change in entropy of the universe due to some block being lifted

  • Thread starter blueyellow
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  • #1
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a 1kg block of steel at temperature 60 degrees is placed at the bottom of 10m deep lake at temprature 10 degrees. the specific heat capacity of steel is 420J/Kkg and is approximately temperature independent.calculate the entropy change of the block, the lake and the universe,after thermal equilibrium has been reached

i did the change in entropy=420 ln ((60+273.16)/(10+273.16))=68.3

so the change in entropy of the block is 68.3, the change in entropy of the lake is zero because the change in temperature of the lake is insignificant and the change in entropy of the universe is the same as the change in entropy of the block.right?

AND

after thermal equilibrium, a crane is used to life the block extremely clowly from the bottom of the lake to the surface, what is the entropy change of the universe?
how am i supposed to know? do i have to take into account the temperature of the surface of the lake/air? we arent told the temperature of the air
 

Answers and Replies

  • #2
supratim1
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i think you are right in the first part.

in 2nd part, i think change in entropy should be zero, as thermal equilibrium is reached.
 
  • #3
Mapes
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so the change in entropy of the block is 68.3, the change in entropy of the lake is zero because the change in temperature of the lake is insignificant
Whoa. (1) How does the block gain entropy if it is cooling down? (2) The change in entropy of the lake isn't negligible, even if the temperature change is.

The block is dumping energy into the lake; we should expect the block's entropy to decrease and the lake's to increase. And we should expect the entropy change of the universe to be positive for any spontaneous process.
 

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