Question

A mass m of lead shot is placed at the bottom of a vertical cardboard cylinder that is and closed at both ends. The cylinder is suddenly inverted so that the shot falls 1.5cm. The process is repeated quickly 100 times, assuming no heat is dissipated or lost, the temperature of the shot will increase by (specific heat of lead =0.03$\dfrac{{cal}}{{{g^0}C}}$ )
A) 0
B) ${5^0}C$
C) ${7.3^0}C$
D) ${11.3^0}C$

Answer 1

Hint:In order to find the solution of this question we need to know about the specific heat which state that number of calories required to raise the temperature of 1 gram of a substance 1°C We have given that the gravitational potential energy is converted into the thermal energy so we will write Q=mgh and equate it to get the solution.

Formula used:
$Q = m \times s \times dT$ Where Q = heat, m = mass, s = specific heat, and dT is the change in temperature ($T_2 - T_1$).

Complete step by step answer:
Step 1:
Before solving the question let us see a definition of specific heat which plays an important role in solving the question.
Specific heat: The number of calories required to raise the temperature of 1 gram of a substance 1°C, or the number of BTU's per pound per degree F. (originally) the ratio of the thermal capacity of a substance to that of standard material.
$Q = m \times s \times dT$ Where Q = heat, m = mass, s = specific heat, and dT is the change in temperature ($T_2 - T_1$). You can use this formula to calculate the specific heat. In the case of gases, just replace the mass by moles of the gas.
Step 2:
We are given:
A mass m of lead shot is placed at the bottom of a vertical cardboard cylinder that is closed at both ends.
The specific heat of lead =0.03$\dfrac{{cal}}{{{g^0}C}}$
The gravitational potential energy is converted into the thermal energy which is absorbed by the lead shot.
If ${T_0}$ is the rise of the temperature in the lead shot then,
$ \rightarrow n\left( {mgh} \right) = \left( {m \times s \times T} \right)J$ Where, n is number of times the process repeated, bracketed part is the potential energy which is m for mass, g for gravity and height as h from which shot is dropped.
Rearranging it to find the temperature and cancelling out mass as common term, $T = \dfrac{{n \times g \times h}}{{s \times J}}$
Substituting the value to find T we get, $T = {\dfrac{{100 \times 9.8 \times 1.5}}{{0.03 \times 4.18 \times {{10}^3}}}^0}C$
On solving this with a simple mathematics we will get value of T=11.3$^0C$
Hence, the temperature of the shot will increase by T=11.3$^0C$

Option D is correct.

Note:The summary of the above story is that the potential energy is getting converted into the thermal energy which is Q then substitutes the value of Q with potential energy and then puts it into the formula of specific heat. Be cautious while solving the mathematical part.