Question

A bicycle is moving at a speed of $36km{h^{ - 1}}$ . Brakes are applied, it stops in $4m$ . If the mass of the cycle is $40kg$ , the temperature of the wheel is risen is [specific heat of wheel $0.25cal/{g^\circ }{C^{ - 1}}$ , mass of the wheel $= 5kg$ ].
A. ${0.19^\circ }C$
B. ${0.47^\circ }C$
C. ${4.7^\circ }C$
D. ${1.9^\circ }C$

Answer 1

To solve this question, first we will rewrite the given information about the question and then we will find the kinetic energy with the help of mass of the wheel and the velocity of the bicycle. And, then we can find the change in temperature.

Complete step by step solution:
Given that:
Velocity of the wheel $= 36km{h^{ - 1}} = 10m{s^{ - 1}}$
mass of the wheel $= 5kg = 5000g$
$1cal = 4.2J$
So, Kinetic Energy produced here:
$= \dfrac{1}{2}m.{v^2} \\\ = \dfrac{1}{2} \times 40kg \times 10m{s^{ - 2}} \\\ = 2000J \\\$
Half of the Kinetic Energy will flow as heat and raise the temperature of the wheel, so the heat energy in raising the temperature of the wheel is $1000J$ .
Now with the specific heat capacity of the wheel $\left( S \right)$ ,the mass of the wheel $(m)$ and the heat energy consumed $(H)$ are known, the change in temperature $(\Delta T)$ can be found out by using the formula:
$H = mS\Delta T$
$\therefore \Delta T = \dfrac{H}{{mS}} \\\ = \dfrac{{1000cal}}{{5000g \times 0.25cal/{g^\circ }C \times 4.2}} \\\ = {0.19^\circ }C \\\$
Hence, the correct option is (A.) ${0.19^\circ }C$

Note:
The specific heat can be characterized likewise for materials that change state or arrangement as the temperature and pressing factor change, as long as the progressions are reversible and steady. Accordingly, for instance, the ideas are determinable for a gas or fluid that separates as the temperature increments, as long as the results of the separation immediately and totally recombine when it drops.