Suppose the graphite target is 0.20 kg and is initially at a temperature of 23â—

Question

Suppose the graphite target is 0.20 kg and is initially at a temperature of 23◦C. Typically 3 × 1013 protons collide with the target each pulse, and each proton has an energy of 120 GeV. Most of the energy goes into the creation of particles. If 0.5% of the energy of the protons in a pulse goes into heating the target, what is the final temperature of the target after one pulse of protons? (In reality, the target has a cooling system to prevent such a temperature increase.) The specific heat of graphite is 710 J/(kgK).

Explanation / Answer

Q = M C dT

0.5/100*3E13*120.0E9*1.6E-19 = 0.2*710*(T-23)

T=43.28 C

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.