# A 0.003 0-kg lead bullet is traveling at a speed of 240 m/s when it embeds in a block of ice at 0°C. If all the heat generated goes into melting ice, what quantity of ice is melted?

## (Lf = 80 kcal/kg, the specific heat of lead = 0.03 kcal/kg⋅°C, and 1 cal = 4.186 J)

Jun 27, 2017

I got $\text{0.26 g}$ although I am not 100 % sure of my method…

#### Explanation:

You know that the bullet carries a kinetic knergy $K$ of:

K=1/2mv^2=1/2("0.0030 kg")("240 m/s")^2 = "86.4 J"

This energy will be used to melt the ice (OK, it is not true because most of it will be used to break it, but anyway...)

We can say that this energy as heat will be used to melt a mass ${m}_{i}$ of ice as in:

$Q = K = {m}_{i} {L}_{f}$

So, knowing that for the energy:

$\text{86.4 J = 0.020 64 kcal}$

we get:

$\text{0.020 64 kcal" = 80m_i " kcal/kg}$

Rearranging:

${m}_{i} = \text{0.020 64 kcal"/"80 kcal/kg" = "0.000 258 kg" = "0.26 g}$

I am not sure about the specific heat of lead that I think could be used if we had the original temperature of the bullet to add to the energy generated and going into melting the ice...