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))
Show Hint
- 0.48 K
- 0.36 K
- 0.14 K
- 0.23 K
The Correct Option is A
Approach Solution - 1
To solve the problem of finding the rise in temperature of the water as it falls from a height of 200 m, we will use the concept of energy conservation. The potential energy lost by water falling is converted entirely into thermal energy, resulting in an increase in water temperature.
Step-by-Step Solution:
- Calculate the potential energy (PE) lost by the water:
The potential energy of an object of mass \(m\) at height \(h\) is given by \(PE = mgh\). In this case, \(g = 10 \, \text{m/s}^2\) and \(h = 200 \, \text{m}\).
- Use the conversion of potential energy to heat energy:
When water falls and its potential energy is converted to heat, it causes an increase in temperature. The heat energy is calculated as \(Q = m \times c \times \Delta T\), where \(c\) is the specific heat capacity of water, \(4200 \, \text{J/(kg K)}\).
- Equate the potential energy to heat energy and solve:
\(mgh = m \times c \times \Delta T\)
Cancel \(m\) from both sides:
\(gh = c \Delta T\)
Solve for \(\Delta T\):
\(\Delta T = \frac{gh}{c}\)
Substitute the known values:
\(\Delta T = \frac{10 \times 200}{4200}\) \(\Delta T = \frac{2000}{4200} = 0.476 \, \text{K}\)
Rounding this to two decimal places, we get \(0.48 \, \text{K}\).
Conclusion: The rise in temperature of the water is approximately \(0.48 \, \text{K}\), which is the correct answer.
Approach Solution -2
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.
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