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 m/s², specific heat of water = 4200 J/(kg K))

<p>To find the rise in temperature of the water as it falls from a height, we start by equating the potential energy loss to the heat gained, assuming no heat is lost from the water in the pool.</p> <p>The potential energy lost by the water can be expressed as:</p> <p>$ mgh $</p> <p>where:</p> <p><p>$ m $ is the mass of water,</p></p> <p><p>$ g $ is the acceleration due to gravity (10 m/s²),</p></p> <p><p>$ h $ is the height (200 m).</p></p> <p>The heat gained by the water when temperature changes is given by:</p> <p>$ ms \Delta T $</p> <p>where:</p> <p><p>$ s $ is the specific heat capacity of water (4200 J/(kg K)),</p></p> <p><p>$ \Delta T $ is the change in temperature.</p></p> <p>Setting the potential energy loss equal to the heat gained:</p> <p>$ mgh = ms \Delta T $</p> <p>We can simplify this equation by canceling out the mass $ m $:</p> <p>$ gh = s \Delta T $</p> <p>Now, solve for $ \Delta T $:</p> <p>$ \Delta T = \frac{gh}{s} $</p> <p>Substitute the known values:</p> <p>$ \Delta T = \frac{10 \times 200}{4200} $</p> <p>Calculate $ \Delta T $:</p> <p>$ \Delta T = \frac{2000}{4200} $</p> <p>Simplify the fraction:</p> <p>$ \Delta T = \frac{10}{21} $</p> <p>Therefore, the rise in temperature is $ \Delta T \approx 0.48 \, \text{K} $.</p>