A hammer of mass 1 kg having speed of 50 text m/s, hit a iron nail of mass 200gm . If specific heat of iron is 0.105cal/gmoC and half the energy is converted into heat, the raise in temperature of nail is:

Question

A hammer of mass 1 kg having speed of 50 text m/s, hit a iron nail of mass 200gm . If specific heat of iron is 0.105cal/gmoC and half the energy is converted into heat, the raise in temperature of nail is: (A)  7.1oC  (B)  9.2oC  (C)  10.5oC  (D)  12.1oC

Tardigrade Admin · Accepted Answer

According to law of conservation heat. ms Δ t=(1/2)( (1/2)m upsilon 2 )× (1/4.2) Δ t=(1× 50× 50/0.2× 4× 4.2× 0.105× 103)=7.1oC

Q. A hammer of mass 1 kg having speed of $50 m / s,$ hit a iron nail of mass $200 g m$ . If specific heat of iron is $0.105 c a l / g m^{o} C$ and half the energy is converted into heat, the raise in temperature of nail is:

$7.1^{o} C$

$9.2^{o} C$

$10.5^{o} C$

$12.1^{o} C$

According to law of conservation heat. $m s Δ t = \frac{1}{2} (\frac{1}{2} m υ^{2}) \times \frac{1}{4.2}$ $Δ t = \frac{1 \times 50 \times 50}{0.2 \times 4 \times 4.2 \times 0.105 \times 10 ^{3}} = 7.1^{o} C$

Q. A hammer of mass 1 kg having speed of 50m/s, hit a iron nail of mass 200gm . If specific heat of iron is 0.105cal/gmoC and half the energy is converted into heat, the raise in temperature of nail is:

7.1oC

9.2oC

10.5oC

12.1oC