Question

A water fall is 84 metres high. If half of the potential energy of the falling water gets converted to heat, the rise in temperature of water will be

• A. 0.098°C
• B. 0.98°C
• C. 9.8°C
• D. 0.0098°C

Answer 1

Answer: (A)

0.098°C

Answer 2

As $W = JQ$ ⇒ $\frac{1}{2}(mgh) = J \times mc\Delta\theta$ ⇒ $\Delta\theta = \frac{gh}{2JS}$

$\Delta\theta = \frac{9.8 \times 84}{2 \times 4.2 \times 1000} = 0.098{^\circ}C$

$(\because S_{\text{water}} = 1000\frac{cal}{kg \times {^\circ}C})$

Short trick : Remember the value of $\frac{g}{Jc_{W}} = 0.0023$,

here $\Delta\theta = \frac{1}{2} \times (0.0023)h = \frac{1}{2} \times 0.0023 \times 84 = 0.098{^\circ}C$