A 2.6 kg rock is initially at rest at the top of a cliff. Assuming the rock falls into the sea at the foot of the cliff and that its kinetic energy is transferred entirely to the water, how high is the cliff if the temperature of 0.98 kg of water is raised 0.10°C? (Neglect the heat capacity of the rock.)

I answered this earlier. See the posts at the bottom.

To determine the height of the cliff, we can use the principle of conservation of energy.

First, we need to calculate the potential energy the rock has at the top of the cliff. The potential energy is given by the equation:

PE = m * g * h

Where:
PE = potential energy
m = mass of the rock (2.6 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the cliff (unknown)

Next, we need to calculate the kinetic energy the rock has when it falls to the foot of the cliff. The kinetic energy is given by the equation:

KE = 0.5 * m * v^2

Where:
KE = kinetic energy
m = mass of the rock (2.6 kg)
v = velocity of the rock (unknown)

According to the problem, all of the rock's kinetic energy is transferred entirely to the water. Therefore, we can equate the kinetic energy to the amount of heat transferred to the water:

KE = m_w * C_w * ΔT

Where:
m_w = mass of water (0.98 kg)
C_w = specific heat capacity of water (4186 J/kg °C)
ΔT = change in temperature of the water (0.10 °C)

In this case, the change in temperature of the water represents the increase in its internal energy, which is equivalent to the rock's kinetic energy.

Now, we can set up two equations by equating the potential energy and kinetic energy:

PE = KE

m * g * h = m_w * C_w * ΔT

Substituting the given values, we have:

2.6 kg * 9.8 m/s^2 * h = 0.98 kg * 4186 J/kg °C * 0.10 °C

Simplifying:

25.48 h = 4087.64

To solve for h, divide both sides by 25.48:

h = 160.3 meters

Therefore, the height of the cliff is approximately 160.3 meters.