Question

A piece of lead falls from a height of $ 100m $ on a fixed non-conducting slab which brings it to rest. If the specific heat of lead is $ 30.6cal/k{g^ \circ }C $ , the increase in temperature of the slab immediately after collision is :
(A) $ {7.62^ \circ }C $
(B) $ {6.72^ \circ }C $
(C) $ {5.62^ \circ }C $
(D) $ {8.72^ \circ }C $

Answer 1

Hint: The lead piece falls from a specified height and has a certain value of potential energy to it while falling under the influence of gravity. Now, the lead piece is said to collide with the slab which is said to be non-conducting. The potential energy experienced by the block is converted to thermal energy. Find the increase in temperature using the equation.

Complete Step by Step Solution
The lead piece is said to experience a free fall motion under gravity from a height of $ 100m $ and lands on a fixed slab. Now, the lead piece is said to be experiencing a potential energy due to free fall under gravity. This can be written as the product of its mass, acceleration due to gravity and the height at which it is dropped from. Mathematically, this is represented as ,
 $ \Rightarrow PE = mgh $ , where m is mass of the object, h is the height from which the object is dropped.
Now, it is said that on collision with a static slab , the molecules of lead undergo movement to produce a change in temperature . The specific heat of the lead particle is given . When the particle hits the non-conducting slab, the potential energy gets converted into thermal energy. Hence, this can be equated as,
 $ \Rightarrow PE = mgh = mS\Delta T $
Cancelling the common mass term, we get,
 $ \Rightarrow gh = S\Delta T $
Taking specific heat to the other side, we get,
 $ \Rightarrow \dfrac{{gh}}{S} = \Delta T $
Substituting the values, we get,
 $ \Rightarrow \dfrac{{(9.81)(100)}}{{30.6cal/k{g^ \circ }C}} = \Delta T $
It is given as cals, now we can convert this to joules. We know that, $ 1Cal = 4.2J $
 $ \Rightarrow \dfrac{{(9.81)(100)}}{{30.6 \times 4.2}} = \Delta T $
Solving this we get,
 $ \Rightarrow \Delta T = {7.62^ \circ }C $
Hence, option (A) is the right answer for the given question.

Note:
Specific heat of a substance is defined as the amount of heat required to raise the temperature of the unit mass of the system by 1 degree. It is measured in $ {J^{}}{g^{ - 1}}{K^{ - 1}} $ or $ ca{l^{}}{g^{ - 1}}{({}^ \circ C)^{ - 1}} $ . The specific heat of a substance is given by the formula
 $ Q = mc\Delta T $ , where c is the specific heat of the substance. Sometimes c is also interpreted as S. Here the formula changes as $ Q = mS\Delta T $ , where S is specific heat of the substance. It is convenient that the student follows any one method.