Thermal Properties of Matter

JEE Main level, single correct, medium

67 of 126

Question 67 of 126

A 100g100g of iron nail is hit by a 1.5kg1.5kg hammer striking at a velocity of 60ms−160{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 =0.42Jg−1C∘−1=0.42{Jg}^{-1}{C^{\circ}}^{-1}]

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Answer: (A) 16.07∘C16.07^{\circ}C

If we heat-up an object, then, the energy is used to either change the temperature of the body or to change the state of the body.
If it changes the temperature, then,
Q=mS∆TQ=mS∆T
where, QQ is the heat given, mm is the mass and ∆T∆T is the change in temperature.
We know that a moving object has an energy in the form of kinetic energy, i.e.,
KE=12mv2KE=\dfrac{1}{2}m{v}^{2}
When that moving object strikes another object, it loses some energy which changes into the heat energy and this heat energy increases the temperature of the object.
Given a hammer of 1.5kg1.5kg is moving with 60ms−160m{s}^{-1}.
Kinetic energy =12mv2=12(1.5)(60)2=2700J=\dfrac{1}{2}m{v}^{2}=\dfrac{1}{2}(1.5)(60{)}^{2}=2700J
One-fourth of kinetic energy changes into heat energy.
⇒14(KE)=\Rightarrow \dfrac{1}{4}(KE)= Heat energy
⇒14(KE)=m(S)(∆T)\Rightarrow \dfrac{1}{4}(KE)=m(S)(∆T)
⇒14(2700)=(100)(0.42)(∆T)\Rightarrow \dfrac{1}{4}(2700)=(100)(0.42)(∆T)
⇒∆T=16.07∘C\Rightarrow ∆T=16.07^{\circ}C

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