Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 14, Problem 97P

(a)

To determine

The increasing temperature of the bullet.

(a)

Expert Solution
Check Mark

Answer to Problem 97P

The increasing temperature of the bullet is 180°C.

Explanation of Solution

Write the equation for heat energy required to produce temperature change in a system.

Q=mcΔT (I)

Here, Q is the heat energy, m is the mass of the iron bullet, c is the specific heat capacity of the iron material, and ΔT is the increasing temperature of the bullet.

Write the equation for kinetic energy of system.

E=12mv2 (II)

Here, E is the kinetic energy of the iron bullet and v is the speed of the bullet.

Conclusion:

Compare the equation (I) and (II) and solve for ΔT.

mcΔT=12mv2cΔT=12v2ΔT=v22c

Substitute 4.00×102m/s for v and 0.44kJ/kgK for c in above equation to find ΔT.

ΔT=(4.00×102m/s)22(0.44kJ/kgK)(103J1kJ)=(4.00×102m/s)22(0.44×103J/kgK)=182°C180°C

Therefore, the increasing temperature of the bullet is 180°C.

(b)

To determine

Find the equilibrium temperature.

(b)

Expert Solution
Check Mark

Answer to Problem 97P

The equilibrium temperature is 20.9°C.

Explanation of Solution

Write the equation for heat energy required to produce temperature change in iron bullet.

QFe=mFecFeΔT (III)

Here, QFe is the heat energy acquired by the bullet, mFe is the mass of the iron bullet, c is the specific heat capacity of the iron material, and ΔT is the increasing temperature of the bullet.

Write the equation for heat energy required to produce temperature change in wooden block.

QW=mWcWΔT (IV)

Here, QW is the heat energy acquired by the wooden block, mW is the mass of the wooden block, cW is the specific heat capacity of the wooden block, and ΔT is the increasing temperature bullet and block system.

Write the equation for kinetic energy of system.

EFe=12mFev2 (V)

Here, EFe is the kinetic energy of the iron bullet and v is the speed of the bullet.

The change in internal energy of the bullet and block system is equal to the initial kinetic energy of the bullet so that ΔU=E.

ΔU=QFe+QW12mFev2=mFecFeΔT+mWcWΔT12mFev2=(mFecFe+mWcW)ΔT

Replace TfTi for ΔT and solve the relation for Tf.

12mFev2=(mFecFe+mWcW)(TfTi)12mFev2+Ti=(mFecFe+mWcW)TfTf=mFev22(mFecFe+mWcW)+Ti (VI)

Conclusion:

Substitute 4.00×102m/s for v, 10.0g for mFe, 0.44kJ/kgK for cFe, 20.0°C for Ti, 0.500kg for mW, and 1680J/kgK for cW in equation (VI) to find Tf.

Tf=(10.0g)(103kg1g)(4.00×102m/s)22[(10.0g)(103kg1g)(0.44kJ/kgK)(103J1kJ)+(0.500kg)(1680J/kgK)]+20.0°C=(10.0×103kg)(4.00×102m/s)22[(10.0×103kg)(0.44×103J/kgK)+(0.500kg)(1680J/kgK)]+20.0°C=1600kgm2/s22[(4.4J/K)+840J/K]+20.0°C=20.9°C

Therefore, the equilibrium temperature is 20.9°C.

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