A horizontal fire hose with a nozzle of cross | vedclass

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(A) Given,cross-sectional area of the nozzle $A = \frac{5}{\sqrt{21}} 	imes 10^{-3} 	ext{ m}^2$.Volume flow rate $Q = \frac{1 	ext{ m}^3}{10 	ext{

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(A) Given,cross-sectional area of the nozzle $A = \frac{5}{\sqrt{21}} 	imes 10^{-3} 	ext{ m}^2$.Volume flow rate $Q = \frac{1 	ext{ m}^3}{10 	ext{ s}} = 0.1 	ext{ m}^3/	ext{s}$.The velocity of water $v$ is given by $v = \frac{Q}{A} = \frac{0.1}{\frac{5}{\sqrt{21}} 	imes 10^{-3}} = \frac{10^{-1} 	imes \sqrt{21}}{5 	imes 10^{-3}} = 20\sqrt{21} 	ext{ m/s}$.When the water hits a rigid wall,its kinetic energy is converted into heat energy. The maximum possible increase in temperature $\Delta T$ is given by the energy balance equation: $\frac{1}{2}mv^2 = ms\Delta T$.Here,$s$ is the specific heat capacity of water $= 4.2 	imes 10^3 	ext{ J/(kg} \cdot ^{\circ}	ext{C)}$.Thus,$\Delta T = \frac{v^2}{2s} = \frac{(20\sqrt{21})^2}{2 	imes 4.2 	imes 10^3}$.$\Delta T = \frac{400 	imes 21}{8.4 	imes 10^3} = \frac{8400}{8400} = 1^{\circ} 	ext{C}$.

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