Question

A hammer of mass $1\text{kg}$ having a speed of $50\text{m/s}$ hits an iron nail of mass $200\text{gm}$. If the specific heat of iron is $0.105{\text{cal/gm}}^{\circ }\text{C}$ and half the energy of the hammer is converted into heat, then the rise in the temperature of the nail is

• A. ${7.1}^{\circ }\text{C}$
• B. ${9.2}^{\circ }\text{C}$
• C. ${10.5}^{\circ }\text{C}$
• D. ${12.1}^{\circ }\text{C}$

Answer 1

The correct option is A ${7.1}^{\circ }\text{C}$

Let us assume,
$W=$ Mechanical work done by hammer converted into heat
$H=$ Heat energy produced
$J=$ mechanical equivalent of heat
$\Delta T=$ change in temperature
We know that $W=JH$
Heat produced in iron nail is given by $H=m\times c\times \Delta T$
$\therefore$ From the data given in the question, we can say that
$\frac{1}{2}\left(\frac{1}{2}M{v}^2\right)=J(m\times c\times \Delta T)$
$\Rightarrow \frac{1}{4}\times 1\times (50{)}^2=4.2[200\times 0.105\times \Delta T]$
$\Rightarrow \Delta T={7.1}^{\circ }\text{C}$
Thus, option (a) is the correct answer.