An iron ball is dropped onto a pavement from a height H. If half of the heat generated goes into warming the ball, find the temperature increase of the ball. Specific heat capacity of iron, Ci = 450 J/kgC H = 183 m. Use acceleration of free fall g = 10 m/s2. Enter 2 digits after decimal point.

Respuesta :

The given question is incomplete. The complete question is as follows.

A 10 kg iron ball is dropped onto a pavement from a height of 100 m. If half of the heat generated goes into warming the ball,find the temperature increase of the ball. (In Sl units, the specific heat capacity of iron is 450 J/kg.C degree).

Explanation:

In the given case, when the ball comes in contact with the pavement then its kinetic energy will be equal to potential energy of the ball before it was dropped. Half the K.E heats the ball.

As K.E = P.E = mgh

So, mgh = [tex]10 \times 10 \times 100[/tex] = 10,000 joules

And, half of this is 5,000 joules

As we know that

             Q = [tex]m \times C \Delta T[/tex]

          5000 = [tex]450 \times 10 \times \Delta T[/tex]

[tex]\Delta T = \frac{5000}{4500}[/tex]

              = 1.11 degrees

Also, in the given situation

          [tex]\frac{mgh}{2} = m \times C \times \Delta T[/tex]

Since there is an m on both sides of the equation, we will divide both sides by m  and the resulting equation will be as follows.

        [tex]\frac{gh}{2} = c \times \Delta t[/tex]

As, the result is independent of mass which is why the result is the same for any ball of any mass.