A hammer of mass 200 kg strikes a steel block of mass 200 g with a velocity 8 ms-1. If 23 % of the energy is utilized to heat the steel block, the rise in temperature of the block is (specific heat capacity of steel = 460 J kg-1 K-1)

Question

A hammer of mass 200 kg strikes a steel block of mass 200 g with a velocity 8 ms-1. If 23 % of the energy is utilized to heat the steel block, the rise in temperature of the block is (specific heat capacity of steel = 460 J kg-1 K-1) (A) 8 K (B) 16 K (C) 12 K (D) 24 K

Tardigrade Admin · Accepted Answer

Given, mass of hammer, m=200 kg, steel block of the mass =200 g =0.2 kg and specific heat capacity of steel, s=460 J kg -1 K -1 Velocity of hammer, v=8 ms -1 As we know that, Kinetic energy, KE =(1/2) m v2 Putting the given values, we get =(1/2) × 200 × 82=6400 J Hence, the 23 % of this energy is converted to heat. ⇒ H=(6400 × 23/100)=1472 J The rise in temperature of steel, Δ T=(H/m s)=(1472/460 × 0.2)=16 K Hence, the rise in temperature is 16 K.

Q. A hammer of mass 200kg strikes a steel block of mass 200g with a velocity 8ms−1. If 23% of the energy is utilized to heat the steel block, the rise in temperature of the block is (specific heat capacity of steel = 460Jkg−1K−1)

8 K

16 K

12 K

24 K

Q. A hammer of mass $200 k g$ strikes a steel block of mass $200 g$ with a velocity $8 m s^{- 1}$ . If $23%$ of the energy is utilized to heat the steel block, the rise in temperature of the block is (specific heat capacity of steel = $460 J k g^{- 1} K^{- 1}$ )