Question

A flashlight battery is hooked to a toy motor, and then the battery and the connections are sprayed with a water-proof coating. The battery is immersed in a beaker holding 100 mL of water. When the toy motor drives a device that raises a weight of 1.00 kg a distance of 10.0 m, the temperature of the water falls by 0.024 C. Assuming that the heat capacity of the battery is negligible compared to that of the water, Calculate the change in the energy of the battery contents as a result of the chemical reaction that took place in the battery.

Answer 1

Explanation:

Formula to calculate work done by motor is as follows.

         Work done by motor = $mass \times g \times height$

where,   g = gravitational constant = 10 $m/s^{2}$

Therefore, work done by motor is as follows.

 Work done by motor = $1.00 kg \times 10 m/s^{2} \times 10.0 m$  

                                  = 100.0 J

Now, heat lost by water will be calculated as follows.

            q = $mC \times \Delta T$

                = $g \times 4.184 J/g^{o}C \times 0.024^{o}C$

                = 10.0 J

Hence, heat gained by motor = heat lost by water

As, heat gained by motor = 10.0 J

So, change in energy = heat gained - work done

Therefore, change in energy will be calculated as follows.  

   Change in energy = heat gained - work done

                                  = (10.0 J) - (100.0 J)

                                   = -90.0 J

Thus, we can conclude that change in the energy of the battery contents is -90.0 J.