What is the temperature increase of water per kilogram at the bottom of a 145 m waterfall if all of the initial potential energy is transferred as heat to the water?
To calculate the temperature increase of water per kilogram at the bottom of a waterfall, we need to use the principle of conservation of energy.
First, let's find the potential energy of the water at the top of the 145 m waterfall. The potential energy (PE) is given by the equation PE = m * g * h, where m is the mass of water, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the waterfall.
Next, we consider that all of the initial potential energy is converted into heat at the bottom of the waterfall. We can use the equation Q = m * c * ΔT, where Q is the heat transferred, m is the mass of water, c is the specific heat capacity of water (approximately 4186 J/kg°C), and ΔT is the change in temperature.
Since all the potential energy is converted into heat, we can equate the potential energy to the heat transferred: m * g * h = m * c * ΔT.
Simplifying the equation, we can cancel out the mass (m) on both sides: g * h = c * ΔT.
Rearranging the equation to solve for ΔT, we have ΔT = (g * h) / c.
Plugging in the values for g (9.8 m/s²), h (145 m), and c (4186 J/kg°C), we can calculate the temperature increase per kilogram at the bottom of the waterfall.
ΔT = (9.8 * 145) / 4186
Calculating this, we find that the temperature increase per kilogram at the bottom of the waterfall is approximately 0.338°C.