Water falls from a height of 200 m. What is the increase in temperature when it touches the bottom? (Assume that all the heat goes into the same amount of mass which was falling).
Show Hint
When an object falls and all of its potential energy converts to heat, use the relationship between energy, mass, and temperature change to find the temperature increase. Remember to use the specific heat capacity for water.
Step 1: Calculate potential energy.
The water's potential energy \( E \) at a height of 200 m is calculated using \( E = mgh \), where \( m \) is mass, \( g = 9.8 \, \text{m/s}^2 \) is gravitational acceleration, and \( h = 200 \, \text{m} \) is height.
Thus, \( E = m \times 9.8 \times 200 = 1960m \, \text{J} \).
Step 2: Determine temperature increase from heat energy.
Assuming all potential energy converts to heat \( Q \) that raises the water's temperature, we use \( Q = mc\Delta T \), where \( c = 4200 \, \text{J/kg}^\circ \text{C} \) is the specific heat capacity of water and \( \Delta T \) is the temperature change.
Equating potential energy to heat energy: \( 1960m = mc\Delta T \).
Simplifying: \( 1960 = 4200 \times \Delta T \).
Solving for temperature change: \( \Delta T = \frac{1960}{4200} = \frac{10}{21}^\circ \, \text{C} \).
Final Answer:
The temperature increase is \( \frac{10}{21}^\circ \, \text{C} \), corresponding to option (2).
