Question

A load $31.4kg$ is suspended from a wire of radius ${10^{ - 3}}m$ and density $9 \times {10^3}kg/{m^3}$. Calculate the change in temperature of the wire if $75\%$ the work done is converted into heat. The young’s modulus and heat capacity of the material of the wire is $9.8 \times {10^{10}}N/{m^2}$ and $490J/kgK$ respectively.

Answer 1

We know about the modulus of rigidity is defined as the ratio of stress and strain. Whenever a load is suspended from a wire there will be an extension in the length of the wire. To explain this young’s modulus is studied, it is the ratio of longitudinal stress and longitudinal strain.

Complete step by step solution:
Let us first write the information given in the question.
Mass $m = 31.4kg$, the radius of wire $r = {10^{ - 3}}m$, density $\rho = 9 \times {10^3}kg/{m^3}$, young’s modulus $Y = 9.8 \times {10^{10}}N/{m^2}$, the heat capacity of the material $C = 490J/kgK$, and $75\%$ of the work is converted into heat.
For any system work done is stored in the form of the potential energy of the system, so we can write work as follows.
$W = \dfrac{1}{2}F \times \Delta l$ ……………………….(1)
Here, $F$ is the force acting on the wire, $\Delta l$ is the change in the length of the wire.
The young’s modulus is given by the following relation.
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}$
Take out the value of $\Delta l$ from the above expression.
$\Delta l = \dfrac{{Fl}}{{YA}}$
Now, substitute this value in equation (1).
$W = \dfrac{1}{2}F \times \dfrac{{Fl}}{{YA}} \Rightarrow W = \dfrac{{l{F^2}}}{{YA}}$ (2)
It is given in the question that 75% of work is converted into heat.
$Q = \dfrac{{75}}{{100}}W$
Substitute the value of work from equation (2).
$Q = \dfrac{{75}}{{100}} \times \dfrac{{l{F^2}}}{{YA}}$
We know, the heat is given below.
$Q = mC\Delta T$
So, we can write the following expression.
$mC\Delta T = \dfrac{{75}}{{100}} \times \dfrac{{l{F^2}}}{{YA}} \Rightarrow \Delta T = \dfrac{{.75 \times l{F^2}}}{{YA\left( {\rho Al} \right)}} = \dfrac{{0.75{F^2}}}{{Y{A^2}\rho }}$
Now, let us put the values in the above formula.
$\Delta T = \dfrac{{0.75{F^2}}}{{Y{A^2}\rho V}} = \dfrac{{0.75 \times {{\left( {31.4 \times 9.8} \right)}^2}}}{{9.8 \times {{10}^{10}} \times {{\left( {\pi {{\left( {{{10}^{ - 3}}} \right)}^2}} \right)}^2} \times 9 \times {{10}^3}}}$
Let us simplify this expression and find the value of the change in the temperature.
$\Delta T = 0.0083$
Hence, the temperature change is $0.008$.

Note:
Heat capacity is the amount of heat required to increase the temperature of the metal by one degree Celsius.
In this question, the length of the wire was not given, so we can calculate or replace it with the following formula.
Volume$=$ area $\times$ length $=$ mass $\times$ density.