Answer :
Answer:
The rise in temperature is [tex]43.98^{\circ}C[/tex]
Solution:
As per the question:
Mass of hammer, M = 1.30 kg
Speed of hammer, v = 7.3 m/s
Mass of iron, [tex]m_{i} = 14\ g[/tex]
No. of blows, n = 8
Specific heat of iron, [tex]C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C[/tex]
Now,
To calculate the temperature rise:
Transfer of energy in a blow = Change in the Kinetic energy
[tex]\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J[/tex]
For 8 such blows:
[tex]\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J[/tex]
Now, we know that:
[tex]Q = m_{i}C_{i}\Delta T[/tex]
[tex]\Delta T= \frac{\Delta KE}{m_{i}C_{i}}[/tex]
[tex]\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C[/tex]