Of two masses of 5 , kg each falling from a h | vedclass

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(D) The potential energy lost by the falling masses is converted into work done to stir the water,which in turn increases the internal energy (tempera

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$0.12$

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(D) The potential energy lost by the falling masses is converted into work done to stir the water,which in turn increases the internal energy (temperature) of the water.Let $m = 5 \, kg$ be the mass of each falling object,$h = 10 \, m$ be the height,$M_w = 2 \, kg$ be the mass of water,and $c = 4200 \, J/kg \cdot ^\circ C$ be the specific heat capacity of water.The total potential energy lost is $PE = 2 	imes mgh = 2 	imes 5 	imes 9.8 	imes 10 = 980 \, J$ (using $g = 9.8 \, m/s^2$).Using the mechanical equivalent of heat,$W = JQ$,where $J = 4.2 \, J/cal$ is often used in older textbook contexts,or simply equating energy in Joules: $PE = M_w c \Delta 	heta$.$980 = 2 	imes 4200 	imes \Delta 	heta$.$\Delta 	heta = \frac{980}{8400} \approx 0.1166 \, ^\circ C$.Rounding to the nearest option provided,$\Delta 	heta \approx 0.12 \, ^\circ C$.

Of two masses of $5 \, kg$ each falling from a height of $10 \, m$,by which $2 \, kg$ of water is stirred. The rise in temperature of the water will be ........ $^\circ C$.

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