Question

In all calorimetry problems: assume there are no heat losses or outside heat sources. Use specific heats and heat of fusion, heat of vaporization, boiling point, etc.

Joe (mass 97.2 kg) is sitting in a small, round, outdoor swimming pool of radius 3.26 m.  Before he got into the tub, the water was 86.3 cm deep,and the water was at Joe's body temperature, 37^o C (so the temperature of the water did not change when got into the tub).  Suppose the temperature of the water suddenly drops to 9.8^o C, and all the energy released is used to project Joe straight upward.  How high will Joe rise from the surface of the Earth, in km?  Assume there is negligible friction.  (Of course, this type of event is forbidden by the second law of thermodynamics.)

NOTE: You cannot assume that g is constant!

Answer 1

energy released=mass of water*specific heat of water*temperature difference

mass of water=voulme*density

=pi*radius^2*depth*density=pi*3.26^2*0.863*1000=28.8135*10^3 kg.

then energy released=28.82135*10^3*4186*(37-9.8)=3281575853.92 J

let height of Joe above surface of earth is H.

then final potential energy-initial potential energy=energy release

==>(-G*M*m/(R+H))-(-G*M*m/R)=3281575853.92

where G=gravitational constant=6.674*10^(-11)

M=mass of earth=5.972*10^24 kg

m=mass of Joe

R=radius of earth=6.371*10^6 m

using all the values:

3.874*10^16*((1/R)-(1/(R+H))=3281575853.92

==>H/(R*(R+H))=8.47*10^(-8)

==>H/(R+H)=0.53965

==>(R+H)/H=1/0.53965=1.853

==>R+H=1.853*H

==>H=R/(1.853-1)=1.1723

==>H=1.1723*6371 km=7468.8 km

hence Joe will be thrown to a height of 7468.8 km