In Joule's experiment, an insulated container contains 20 kg of water initially at 25°C. It is stirred by an agitator, which is made to turn by a slowly falling body weighing 40 kg through a height of 4 m. The process is repeated 500 times. The acceleration due to gravity is 9.8 ms-2 . Neglecting the heat capacity of agitator, the temperature of water (in °C) is
A)40.5
B)34.4
C)26.8
D)25

Question

Quizplus · Accepted Answer

The energy transferred to the water is equal to the potential energy lost by the falling body, which is given by $mgh$, where $m$ is the mass of the body, $g$ is the acceleration due to gravity, and $h$ is the height. For 500 falls, the total energy transferred is $500 \times 40 \times 9.8 \times 4$ joules. The specific heat capacity of water is $4.18 \, \text{J/g°C}$, and since there are 20 kg of water, this is $20,000 \, \text{g}$. The temperature change, $\Delta T$, can be found from the equation $Q = mc\Delta T$, where $Q$ is the heat added, $m$ is the mass of the water, and $c$ is the specific heat capacity. Solving for $\Delta T$ gives a temperature increase, and adding this to the initial temperature of 25°C gives the final temperature, which is approximately 34.4°C.

In Joule's Experiment, an Insulated Container Contains 20 Kg of Water