Physics problem worth 10 pts, do you have what it takes?explain answers?

Question

While painting the top of an antenna with a height of 218 m, a worker accidentally lets a 1.00 kg bottle of water fall from his lunchbox. The bottle lands in some bushes at ground level and does not break.

If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature in degrees C?
Use 9.81 m/s^2 for the acceleration due to gravity and 4190 J/(kg*K) for the heat capacity at constant pressure for water.

Answer 1

To solve this problem, one assumes that all of the gravitational Potential energy possesed by the water when resting atop the 218 meter tall antenna is converted into heat energy which then goes into raising the temperature of the water.

The Potential Energy the bottle contains is given by the equation,
PE = mgh
where m is the mass, g is the acceleration due to gravity, and h is the height above the ground of the bottle.
PE = (1 kg)(9.81 m/s^2)(218 m)
PE = 2138.58 Joules

To find the change in temperature, we use this formula,
Q = mc(delta T)
where m is the mass of the water, c is the water's specific heat, and T is the change in temperature experienced by the water.

We assume Q = PE, m is a constant 1 kg, and c is given as 4190 J/kg * K,
(2138.58 Joules) = (1 kg)(4190 J/kg * K) [delta T]
delta T = 2138.58 / 4190
delta T = .51 Kelvin

Thus, the 1 kg of water's temperature is calculated to increase by about .5 degrees C after falling from a height of 218 meters assuming a 100% conversion from PE to Q.

Answer 2

Total Mechanical Energy in the bottle =Potential energy @ the top =Kinetic energy @ bottom=Sum of K.E.+ P.E. at any point in between the fall


It is easiest to calculate the P.E. at the top=mgh=1*9.8*218= 2136.4

This will go into the water in the end hence,
2136.4= 1*4190 * (change in temp)

====> Change in temp = 2136.4/4190=0.509 degree C


I am assumiong here that the mass of bottle is negligible ...as nothing else is given in the prob statement

Answer 3

TMEtop=TMEbottom

PEtop+KEtop=PEbottom+KEbottom

KEtop=PEbottom=0

PEtop=mgh=KEbottom

mgh=(9.8 m/s^2)(1.00 kg)(218 m)

PEtop=KEbottom=change in PE=2200 J (sig figs conserved)
****************
Q=mc(delta)T

Q=2200 J
m=1.00 kg
c=4190 J/kg*K

deltaT=.525 C