Water falls from a height of $200 \text{ m}$ 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 $= 4200 \text{ J/(kg K)}$)
Physics Principle: Energy Conservation
The process involves the conversion of potential energy into internal energy (heat). When water falls from a height, this potential energy is transformed. Assuming no heat is lost to the surroundings (no heat dissipation), the gained heat energy directly increases the water's temperature.
The core principle is that the potential energy lost by the water equals the heat energy it gains.
Energy Calculation Steps
Potential Energy (PE) lost by mass $m$ falling height $h$: $PE = mgh$ Where $g$ is acceleration due to gravity ($10 \text{ m/s}^2$) and $h$ is the height ($200 \text{ m}$).
Heat Energy (Q) gained by mass $m$ causing temperature rise $\Delta T$: $Q = mc\Delta T$ Where $c$ is the specific heat of water ($4200 \text{ J/(kg K)}$).
Temperature Rise Calculation
By the conservation of energy (PE lost = Heat gained): $mgh = mc\Delta T$
Mass $m$ cancels out, leaving:
$gh = c\Delta T$
Rearranging to find $\Delta T$:
$\Delta T = \frac{gh}{c}$
Calculation Details
Substitute the given values:
$g = 10 \text{ m/s}^2$
$h = 200 \text{ m}$
$c = 4200 \text{ J/(kg K)}$
$\Delta T = \frac{(10 \text{ m/s}^2) \times (200 \text{ m})}{4200 \text{ J/(kg K)}}$
$\Delta T = \frac{2000 \text{ J/kg}}{4200 \text{ J/(kg K)}}$
$\Delta T = \frac{20}{42} \text{ K} = \frac{10}{21} \text{ K}$
Calculating the final value:
$\Delta T \approx 0.476 \text{ K}$
Rounded to two decimal places, $\Delta T = \boldsymbol{0.48 \text{ K}}$.
Final Temperature Rise
The calculated rise in water temperature is approximately $0.48 \text{ K}$.
Similar Questions
During the melting of a slab of ice at $273 \ K$ at atmospheric pressure:
- $\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right)$, then the value of n is _________.
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Important Questions from Heat and Thermodynamics
During the melting of a slab of ice at $273 \ K$ at atmospheric pressure:
- $\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right)$, then the value of n is _________.
- Match List - I with List - II.$\Delta Q$=Heat supplied
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$\Delta V$ = Change in volume of the system
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Match the LIST-I with LIST-II Choose the correct answer from the options given below:
