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a sophomore with nothing better to do adds heat to 0.350 kg of ice at 0.0oc until it is all melted. a) what is the change in entropy of the water? b) the source of heat is a very massive body at a temperature of 25.0oc. what is the change in entropy of this body? c) what is the total change in entropy of the water and the heat source?



Answer :

A sophomore with nothing better to do adds heat to 0.350 kg of ice at 0.0oc until it is all melted  change in entropy in water is 428 j/k,massive body is -392J/K and heat source is 35.9J/K

The expression for entropy in terms of change in heat energy and temperature difference is

ds=dQ/T

(a)

Heat is added to ice at 0.0 .C until it melts.

Change in entropy of the water is

ds1=dQ/T

=mL/T

=0.35*3.34*10^5/273.15

=428 J/K

Thus, the change in entropy of the water is 428J/K

(b)

Heat is removed from massive body.

Change in entropy of the massive body is

ds, =0.350x3.34x10*

(0.C=273.15K)

273.15+25.0

=-392 J/K

Thus, the change in entropy of the massive body is -392J/K

(c)

The total change in entropy of the water and the heat source is

ΔStotal = ds, + ds,

= 428-392

=35.9J/K

Thus, the change in entropy of the water and the heat source is 35.9J/K

Learn more about entropy here:

https://brainly.com/question/13999732

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