ANS:B - 34.4

In Joule's experiment, mechanical work is done on the water by the agitator, which transfers energy into the water and increases its temperature. The increase in temperature can be calculated using the principle of conservation of energy. The work done by the falling body in each cycle is given by work=force×distancework=force×distance. The force applied by the falling body is the gravitational force acting on it, given by force=mass×acceleration due to gravityforce=mass×acceleration due to gravity. Therefore, the work done in each cycle is work=mass×acceleration due to gravity×distancework=mass×acceleration due to gravity×distance. Given:

• Mass of falling body = 40 kg
• Height through which the falling body falls = 4 m
• Acceleration due to gravity = 9.8 m/s29.8m/s2

The work done in each cycle is: work=40 kg×9.8 m/s2×4 m=1568 Jwork=40kg×9.8m/s2×4m=1568J This work done on the water in each cycle will increase its internal energy, resulting in a temperature rise. To calculate the temperature rise, we can use the specific heat capacity formula: Q=mcΔT Where:

• Q is the heat added to the water
• m is the mass of the water (20 kg)
• c is the specific heat capacity of water (approximately 4186 J/kg⋅K4186J/kg⋅K)
• ΔT is the change in temperature

The total heat added to the water after 500 cycles is =1568 J/cycle×500 cycles=784000 JQ=1568J/cycle×500cycles=784000J. Using the formula ΔT=mcQ​, we can calculate the change in temperature: 784000 J20 kg×4186 J/kg⋅KΔT=20kg×4186J/kg⋅K784000J​ ΔT≈9.4K The initial temperature of the water is 25 °C25°C, so the final temperature is 25 °C+9.4 K=34.4 °C25°C+9.4K=34.4°C. Therefore, the temperature of the water is approximately 34.4°C. So, the correct option is: 34.4°C