Question

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).

• A. \( 0.7^\circ \, \text{C} \)
• B. \( \frac{10}{21}^\circ \, \text{C} \)
• C. \( \frac{20}{21}^\circ \, \text{C} \)
• D. \( \frac{11}{10}^\circ \, \text{C} \)

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.

Answer 1

The Correct Option is B

Step 1: Use the potential energy formula.
The water falls from a height of 200 m.
The potential energy \( E \) of the water at the top is given by: \[ E = mgh \] where: - \( m \) is the mass of the water, - \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)), - \( h \) is the height from which the water falls (200 m). Thus: \[ E = m \times 9.8 \times 200 = 1960m \, \text{J} \]
Step 2: Heat energy.
We assume all this potential energy converts to heat energy, which raises the temperature of the water.
The heat energy \( Q \) required to increase the temperature of a substance is given by: \[ Q = mc\Delta T \] where: - \( c \) is the specific heat capacity of water (\( 4200 \, \text{J/kg}^\circ \text{C} \)), - \( \Delta T \) is the temperature change. Substitute the values: \[ 1960m = mc\Delta T \] \[ 1960 = 4200 \times \Delta T \] \[ \Delta T = \frac{1960}{4200} = \frac{10}{21}^\circ \, \text{C} \]
Final Answer:
The increase in temperature is \( \frac{10}{21}^\circ \, \text{C} \), corresponding to option (2).