Answer for #7 is B and #8 is A I need solution, thanks! 7) A 10-g lead bullet is

Question

Answer for #7 is B and #8 is A

I need solution, thanks!

7) A 10-g lead bullet is traveling at 1000 m/s and has a temperature of 300º C immediately before it is embedded into a 500-g block of clay at 30º C, where it comes to a complete stop. The heat capacity of the lead bullet = 1.28 J/K, the heat capacity of the clay = 690.5 J/K. Assuming all of the bullet's kinetic energy is turned into thermal energy, what will the temperature of the system be when it reaches equilibrium?

8) Compare the change in entropy of the bullet (Sbullet) to the change in entropy of the clay (dSclay).

The next two problems are related: 7. A 10-g lead bullet is traveling at 1000 m/s and has a temperature of 300° C immediately before it is embedded into a 500-g block of clay at 30° C, where it comes to a complete stop. The heat capacity of the lead bullet 1.28 J/K, the heat capacity of the clay 690.5 J/K. Assuming all of the bullet's kinetic energy is turned into thermal energy, what will the temperature of the system be when it reaches equilibrium? a. 30.5o C b. 37.7°C c. 29.70 C d. 303° C e. 278° C 8. Compare the change in entropy of the bullet (ASbullel) to the change in entropy of the clay (ASclay) c. ASbulilel> ASelay

Explanation / Answer

7) KE of bullet = 0.5*mv2= 0.5*0.01*10002 = 5000 J

This energy completely changes into the thermal energy of the bullet and the clay, hence Q=5000 J

let the equilibrium temperature be T, Tbullet= 300+273 = 573 K, Tclay=30+273 = 303 K

Q= 1.28*(573-T) + 690.5(T-303)

5000=1.28*(573-T) + 690.5(T-303)

Solving the above equation for T we get T=310.7 K or T=310.7-273 = 37.7 C

8) S=Q/T

Sbullet= 1.28*(573-T) / T = 1.28*262.3 /310.7 = 1.080 J/K

Sclay = 690.5(T-303) / T = 690.5*7.7 / 310.7 = 17.1 J/K

Sbullet < Sclay

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