Question

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take $ g = 10 \, \text{m/s}^2 $, specific heat of water = 4200 J/(kg K))

• A. 0.48 K
• B. 0.36 K
• C. 0.14 K
• D. 0.23 K

Hint:

Key concepts for energy-temperature problems: - Potential energy converts to thermal energy: \( mgh = mc\Delta T \) - Specific heat capacity \( c \) determines temperature rise - Always check units consistency (J, kg, K)

Answer 1

The Correct Option is A

Step 1: Calculate the potential energy converted to heat: 
When water falls, its potential energy is converted to kinetic energy and then to thermal energy upon impact.
The potential energy per unit mass is: \[ PE = mgh \] where: \begin{itemize} \item \( m \) = mass of water (1 kg for calculation) \item \( g = 10 \, \text{m/s}^2 \) \item \( h = 200 \, \text{m} \) \end{itemize} \[ PE = 1 \times 10 \times 200 = 2000 \, \text{J} \] 
Step 2: Relate energy to temperature change: 
The thermal energy \( Q \) is related to temperature change \( \Delta T \) by: \[ Q = mc\Delta T \] where: \begin{itemize} \item \( c = 4200 \, \text{J/(kg K)} \) (specific heat capacity of water) \end{itemize} Rearranging for \( \Delta T \): \[ \Delta T = \frac{Q}{mc} = \frac{2000}{1 \times 4200} \approx 0.476 \, \text{K} \] 
Step 3: Round to match options: 
The closest option to 0.476 K is 0.48 K.