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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