A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms -1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron =0.42 Jg -1 ° C -1 ]

Question

A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms -1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron =0.42 Jg -1 ° C -1 ] (A) 675° C (B) 1600° C (C) 160.7° C (D) 6.75° C

Tardigrade Admin · Accepted Answer

beginarrayl text Heatenergy =(1/4) × text Kinetic energy m c Δ T=(1/4) × (1/2) m v2 0.1 × 420 × Δ T=(1/8) × 1.5 ×(60)2 Δ T=(1/8) × (1.5 ×(60)2/0.1 × 420) Δ T=(5400/336) Δ T=16.07° C endarray

Q. A $100 g$ of iron nail is hit by a $1.5 k g$ hammer striking at a velocity of $60 m s^{- 1}$ . What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron $= 0.42 J g^{- 1}^{\circ} C^{- 1}$ ]

$67 5^{\circ} C$

$160 0^{\circ} C$

$160. 7^{\circ} C$

$6.7 5^{\circ} C$

$Heatenergy = \frac{1}{4} \times Kinetic energy m c Δ T = \frac{1}{4} \times \frac{1}{2} m v^{2} 0.1 \times 420 \times Δ T = \frac{1}{8} \times 1.5 \times (60)^{2} Δ T = \frac{1}{8} \times \frac{1.5 \times ( 60 ) ^{2}}{0.1 \times 420} Δ T = \frac{5400}{336} Δ T = 16.0 7^{\circ} C$

Q. A 100g of iron nail is hit by a 1.5kg hammer striking at a velocity of 60ms−1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron =0.42Jg−1∘C−1 ]

675∘C

1600∘C

160.7∘C

6.75∘C