Temperature change in a nail it's a tricky one.

AI Thread Summary
Pounding a nail into wood generates heat, raising the nail's temperature. For a 5-gram steel nail, 6 cm long, with a hammer force of 600 N, the calculation involves work done (W = Fd) and heat transfer (Q = mcΔT). The work done is calculated as 36 J, leading to a temperature increase (ΔT) of 16°C when using the specific heat capacity of steel. A minor typo was noted in the initial work, but the calculations were confirmed to be correct. The discussion effectively demonstrates the relationship between mechanical work and thermal energy in this context.
Dmitri10
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"Pounding an nail into wood makes the nail warmer. Consider a 5-g steel nail 6 cm long, and a hammer that exerts an average force of 600 N on it when driving the nail into a piece of wood. About how much hotter will the nail become? (The specific heat capacity of steel is 450 J/kgC.)"

Here's my work:
Given: m = 0.005 kg, d = 0.06 m, F = 600 N, c = 450 J/kgoC
Unknown: ΔT
Equation: W = Fd, Q = mcΔT
Substitution: W = 600 N * 0.06 W = 30 J
30 J = 0.005 kg * 450 J/kgoC * ΔT
Solution: ΔT = 16 oC

Is this correct?
 
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Dmitri10 said:
"Pounding an nail into wood makes the nail warmer. Consider a 5-g steel nail 6 cm long, and a hammer that exerts an average force of 600 N on it when driving the nail into a piece of wood. About how much hotter will the nail become? (The specific heat capacity of steel is 450 J/kgC.)"

Here's my work:
Given: m = 0.005 kg, d = 0.06 m, F = 600 N, c = 450 J/kgoC
Unknown: ΔT
Equation: W = Fd, Q = mcΔT
Substitution: W = 600 N * 0.06 W = 30 J
30 J = 0.005 kg * 450 J/kgoC * ΔT
Solution: ΔT = 16 oC

Is this correct?
It does look correct, even with your typo of 30 J instead of 36 J, but you used 36 J anyway in your calc, looks good!:wink:
 
Oh! Yes, 30 J was a typo; I actually used 36 J. Thank you for checking my work though!
 
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