Question

In a Joule experiment, a mass of 6.51 kg falls through a height of 66.8 m and rotates a paddle wheel that stirs 0.68 kg of water. The water is initially at 15 degrees C. By how much does its temperature rise (in degrees C)

Answer 1

Explanation:

As the given data is as follows.

         m = 6.51 kg,   g = 9.8 $m/s^{2}$,     h = 66.8 m

mass of water (M) = 0.68 kg,    Specific heat of water = 4200 $J/kg^{o}C$

    $T_{1} = 15^{o}C$

According to the given situation, the decrease in potential energy of mass will be equal to heat energy gained by water.

Therefore,

                mgh = $m \times C \times (T_{f} - T_{i})$

    $6.51 kg \times 9.8 \times 66.8 m = 0.68 kg \times 4200 J/kg^{o}C \times (T_{f} - 15)^{o}C$

        4261.7064 = $2856 (T_{f} - 15)^{o}C$

              1.492 = $T_{f}$ - 15

              $T_{f} = 16.492^{o}C$

So,    $\Delta T = (16.492 - 15)^{o}C$

                       = $1.492^{o}C$

Therefore, we can conclude that rise in temperature will be $1.492^{o}C$.