Question

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

• A. ${0.19}^{o}C$
• B. ${0.47}^{o}C$
• C. ${4.7}^{o}C$
• D. ${1.9}^{o}C$

Answer 1

The correct option is A ${0.19}^{o}C$

Velocity of the wheel=$36km{h}^{-1}$ =$10m{s}^{-1}$
mass of the wheel =$5kg$ =$5000g$
$1cal=4.2J$
Kinetic energy produced = $\frac{1}{2}\times m\times v^2$ =$\frac{1}{2}\times 40kg\times 10m{s}^{-2}$ =$2000J$

Half of this 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($S$),the mass of the wheel($m$) and the heat energy consumed($H$) are known, the change in temperature ($△T$) can be found out by using the formula
$H=mS△T$
$△T=\frac{H}{mS}$
$=\frac{1000cal}{5000g\times 0.25cal/g^{o}C\times 4.2}$
$={0.19}^{o}C$.