A circus stunt diver whose mass is 60.0kg belly flops into a tub containing 8000
ID: 1466675 • Letter: A
Question
A circus stunt diver whose mass is 60.0kg belly flops into a tub containing 8000kg of water from 5.00 m above the surface of water.
a. How much does the tempeture of the water increase due to the diver stopping quickly in the water? Assume all the water stays in the tub, as does the energy trasffered from hte belly flop. Use 4.186 J/(g K) for the specific heat of water.
b. If the diver stays in the tub for 60 sec and loses 100 J per second of heat to the water, how much further does the water temperature increase? Assume no heat trasfer through the walls and the floor of the tub and surface of water.
Explanation / Answer
a)
mass of the diver md=60 kg
mass of the water in tub mw=8000 kg
height h=5m
by energy relation,
md*g*h=mw*Cw*dT
60*9.8*5=8000*4186*dT
===> dT=8.78*10^-5 K
increase in temperature dT=8.78*10^-5 K
b)
here,
energy transfered to the water is,
Q=P*t
Q=100*60
Q=6000 J
this energy is used to increse the temperature of the water
6000=8000*4186*(dT'-8.78*10^-5)
===> dT'=2.67*10^-4 K
increase in temperature dT'=2.67*10^-4 K
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