a tourist drops 1-liter water filled baloon from the top of the Washington monument (170 m high).If air resistance is negligible and almost all the motion energy heats the water,find the increase in temperature of water when it hits the pavement?
mgh=m*c*deltaTEmp
solve for delta temp
To find the increase in temperature of the water when it hits the pavement, we can use the principle of conservation of energy.
The potential energy of the water at the top of the Washington monument is given by:
PE = mgh
Where:
m = mass of the water = 1 kg (since the density of water is close to 1000 kg/m^3 and the volume is 1 liter)
g = acceleration due to gravity = 9.8 m/s^2
h = height of the Washington monument = 170 m
PE = (1 kg) x (9.8 m/s^2) x (170 m) = 1666 J
All of this potential energy will be converted into heat energy when the water hits the pavement (assuming negligible air resistance), so the increase in temperature can be calculated using the specific heat capacity of water:
Q = mcΔT
Where:
Q = heat energy gained by the water
m = mass of the water = 1 kg
c = specific heat capacity of water = 4186 J/kg°C (approximate value)
ΔT = change in temperature of the water
Given that Q = 1666 J, m = 1 kg, and c = 4186 J/kg°C, we can solve for ΔT:
1666 J = (1 kg) x (4186 J/kg°C) x ΔT
ΔT = 1666 J / (1 kg x 4186 J/kg°C)
ΔT ≈ 0.398°C
Therefore, the increase in temperature of the water when it hits the pavement will be approximately 0.398°C.