What was the initial temperature of the lead bullet before it melted completely?

[Vitalius6189]

Homework Statement

A lead bullet flies at a speed of 450 m / s and, striking a wall perpendicularly, melts
completely. What was the bullet temperature before the blow, if for melting it consumes only half of its mechanical energy? Latent melting heat and the specific and melting temperature of the lead are λt
= 25 kJ / kg, c = 130 J / (kg · K) Tmelting= 600 K.

Homework Equations

½ x ½mv² = mcΔT + mλ

The Attempt at a Solution

The bullet (mass m) is first be heated to its melting temperature. Call the temperature-increase ΔT. This requires energy = mcΔT.

The bullet is then melted which requires energy = mλ

The bullet's initial kinetic (mechanical) energy was ½mv². Half of this heats and melts the bullet so:
½ x ½mv² = mcΔT + mλ

v²/4 = cΔT + λ

λ = 25000 J/kg
c = 130 J/(kg·K)

450²/4 = 130ΔT + 25000
ΔT = (67500 - 25000)/130
. . . = 327K

Therefore the temperature has risen 327K to reach melting point (600K) so the initial temperature was 600 - 327 = 273K (i.e. 0°C).
The problem is that the answer is supossed to be aproximately 400 K but i can't get it.
I feel like I'm missing something but don't know what.

Any help will be appreciated.

 

[abilolado]

I think the problem is just the math in the end.
Everything is right up to:
\frac{450^2}{4}\text{=130$\Delta $T+25000}
but \frac{450^2}{4}=50625, not 67500 as you put in your solution.
So then:
\text{$\Delta $T=}\frac{(50625-25000)}{130}\text{=197.115}
So T=600-197.115=402.885K

 

[gneill]

450²/4 = 130ΔT + 25000
ΔT = (67500 - 25000)/130
. . . = 327K

Looks like it may be a calculator problem. Try that calculation again.

 

[Vitalius6189]

I shoulb be way more attentive. Sorry for bothering and thank you for help.