Transforming kinetic energy into thermal energy

AI Thread Summary
In a head-on collision of two cars traveling at 80 km/h, all kinetic energy is converted into thermal energy. The calculations show that the temperature increase for each car is approximately 7.13 K, which translates to 0.55 °C after unit conversion. The specific heat used in the calculation is that of iron, set at 449 J/kg·K. The user clarified the unit conversions, confirming that kinetic energy (J) can be expressed in terms of kg, m, and s, leading to the correct temperature change. The discussion emphasizes the importance of unit consistency in physics calculations.
nlingraham
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Homework Statement



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

Homework Equations



K=1/2mv2

Q=mcΔT

1/2mv2=mcΔT

∴ ΔT=v2/2c

The Attempt at a Solution



I worked it out using 80 km/h for v and 449 J/kg * K for c and got 7.13 K. My question is if this is right, and if it is can someone explain to me how the units cancel out to give just K?
 
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Express J in terms of kg, m and s. What is the dimension of c then?
 
Ok, I got it. J=kg*m2/s2. So that meant I had to convert 80 km/h to 22.22 m/s for everything to cancel out. I ended up getting .55 C. Thank you for your help.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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