What is the Temperature Increase of Cars Colliding at 80 km/hr?

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SUMMARY

The temperature increase of two cars colliding head-on at 80 km/hr, assuming all kinetic energy is converted to thermal energy, can be calculated using the equations for kinetic energy and heat transfer. The correct specific heat for iron is 0.449 J/(g*K), which must be converted to J/(kg*K) for accurate calculations. The final temperature increase (delta T) is determined to be 0.549 K after correcting the specific heat unit. This highlights the importance of unit consistency in thermodynamic calculations.

PREREQUISITES
  • Understanding of kinetic energy calculations (K = 0.5mv^2)
  • Knowledge of heat transfer equations (Q = cm(delta T))
  • Familiarity with specific heat capacity, particularly for iron
  • Ability to convert units between grams and kilograms
NEXT STEPS
  • Study the principles of kinetic energy in collisions
  • Learn about heat transfer and specific heat capacity in thermodynamics
  • Explore unit conversion techniques in physics calculations
  • Investigate real-world applications of energy transformation in vehicle collisions
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and thermodynamics, as well as educators looking for practical examples of energy conversion in collisions.

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Homework Statement



Two cars collide head on while each is traveling at 80 km/hr. Suppose all their kinetic energy is transformed into the thermal energy of the wrecks. What is the temperature increase of each car? (Assume each car's specific heat is that of iron)

Homework Equations



K = .5mv^2
Q = cm(delta T)
iron specific heat = .449 J/(g*K)

The Attempt at a Solution



K = 2(.5m(22.2)^2) = 492.8m

(converted 80 km/hr to m/s)

Q = 492.8m = (.449)(2m)(delta T)
delta T = 548.8 KWhich is the wrong answer. I don't know what i am doing wrong. Can someone please help me. Thanks
 
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Got, it. Just needed to convert iron specific heat from J/(g*K) to J/(kg*K), so delta T is just .549
 

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