A 2.8 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 1.04 kg of water is raised 0.10°C? (Neglect the heat capacity of the rock.)
Answer in m
The rock weighing 2.8 kg is initially at rest at the top of a cliff. It falls into the sea at the foot of the cliff and its kinetic energy is transferred entirely to the water. The height of the cliff has to be determined if the temperature of 1.04 kg of water is raised by 0.10 C.
Let the height of the cliff be H. At the top of the cliff the rock has a potential energy equal to 2.8*9.8*H. When the rock falls the potential energy is converted to kinetic energy and by the time the rock comes to the bottom of the cliff all the potential energy it had has been converted to kinetic energy. The rock falls into 1.04 kg of water raising its temperature by 0.1 C. The heat capacity of water is 4186 J/kg*C. To raise the temperature of 1.04 kg of water by 0.1 C, the energy required is 4186*1.04*0.1 = 435.3 J
This is the kinetic energy of the rock which is also equal to 2.8*9.8*H.
2.8*9.8*H = 435.5
=> H = 15.86
The height of the cliff is 15.86 m.