Question

An iron block of mass $ {\text{2kg}} $ , falls from a height of $ 10{\text{m}} $ . After colliding with the ground it loses $ 25\% $ energy to surroundings and rest is gained as heat. Then find the temperature rise of the block. (Take sp. heat of iron $ 470{\text{ J/k}}{{\text{g}}^ \circ }{\text{C}} $ ).
(A) $ {0.53^ \circ }{\text{C}} $
(B) $ {0.053^ \circ }{\text{C}} $
(C) $ {0.159^ \circ }{\text{C}} $
(D) $ {0.212^ \circ }{\text{C}} $

Answer 1

Hint : To solve this question, we need to find out the potential energy of the block. This energy will be the total energy of the block part of which is gained as heat, which can be found out. The temperature rise of the block can be evaluated by using the value of specific heat.

Formula Used: The formulae which are used to solve this question are given by
 $\Rightarrow P = mgh $ , here $ P $ is the potential energy of a body of mass $ m $ and at a height of $ h $ , $ g $ is the acceleration due to gravity.
 $\Rightarrow Q = ms\Delta T $ , here $ Q $ is the heat given to a body of mass $ m $ and specific heat $ s $ for a temperature rise of $ \Delta T $ .

Complete step by step answer
Let the temperature rise of the block be $ \Delta T $ .
We know that the potential energy of a body which is situated at a height is given by
 $\Rightarrow P = mgh $
According to the question, we have $ m = 2kg $ and $ h = 10m $ . So the potential energy of the block is
 $\Rightarrow P = 2 \times 10 \times 10 = 200J $
Now, as the iron block falls below, its potential energy starts getting converted into kinetic energy. When it hits the ground, its potential energy becomes zero while the kinetic energy becomes maximum and is equal to the initial potential energy. So the kinetic energy at the bottom-most point is given by
 $\Rightarrow K = 200J $ ................................(i)
Now, according to the question $ 25\% $ of this energy is lost to the surroundings and the rest $ 75\% $ of the energy is gained as heat. So the heat gained is given by
 $\Rightarrow Q = 0.75K $
From (i)
 $\Rightarrow Q = 0.75 \times 200J $
 $ \Rightarrow Q = 150J $ …………………………..(ii)
Now we know that the heat required for the temperature rise is given by
 $\Rightarrow Q = ms\Delta T $
According to the question, we have $ m = 2kg $ and $ s = 470{\text{ J/k}}{{\text{g}}^ \circ }{\text{C}} $ . So we have
 $\Rightarrow Q = 2 \times 470 \times \Delta T $
 $ \Rightarrow Q = 940\Delta T $ ………………………….(iii)
Equating (ii) and (ii) we get
 $\Rightarrow 940\Delta T = 150 $
 $ \Rightarrow \Delta T = \dfrac{{150}}{{940}} $
On solving we get
 $\Rightarrow \Delta T = {0.159^ \circ }C $
Thus the temperature rise of the block is equal to $ {0.159^ \circ }{\text{C}} $ .
Hence the correct answer is option C.

Note
This question can be attempted even without using the value of the mass of the iron block, This is because while equating the mechanical energy with the heat, the mass would get cancelled from both the sides of the equation.