Gujarati
10-1.Thermometry, Thermal Expansion and Calorimetry
medium

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

$0.0098°C$

Solution

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

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

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

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

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

Standard 11
Physics

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