A 5.0 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 1.0 kg of water is raised 0.20°C? (Neglect the heat capacity of the rock.)
To solve this problem, we can use the principle of conservation of energy. The potential energy of the rock at the top of the cliff is converted into kinetic energy as it falls, and then converted into thermal energy in the water.
1. First, let's calculate the potential energy of the rock at the top of the cliff using the formula:
Potential energy (PE) = mass × gravity × height
PE = 5.0 kg × 9.8 m/s² × height
2. Next, let's calculate the kinetic energy of the rock just before it hits the water using the formula:
Kinetic energy (KE) = 0.5 × mass × velocity^2
Since the rock starts from rest, the velocity will be the final velocity just before hitting the water.
3. Since all of the rock's kinetic energy is transferred to the water, we can equate the kinetic energy to the heat gained by the water using the formula:
Heat (Q) = mass × specific heat capacity × temperature change
In this case, the mass of water is 1.0 kg, the specific heat capacity of water is 4186 J/kg°C, and the temperature change is 0.20°C.
4. Finally, we can equate the potential energy to the heat gained by the water and solve for the height of the cliff.
5.0 kg × 9.8 m/s² × height = 0.5 × 5.0 kg × velocity^2 = 1.0 kg × 4186 J/kg°C × 0.20°C
Simplifying and rearranging the equation will give you the height of the cliff:
height = (1.0 kg × 4186 J/kg°C × 0.20°C) / (5.0 kg × 9.8 m/s²)