So you're just dividing numbers willy-nilly and disregarding units? Are calories the same as joules?Erenjaeger said:Homework Statement
A 3.20 g lead bullet at 28.0°C is fired at a speed of 210 m/s into a large block of ice at 0°C, in which it becomes embedded. What mass of ice melts (g)?
[/B]Homework Equations
I thought that the KE of the bullet would be transferred into heat [/B]The Attempt at a Solution
I used MV^2/2 = 0.0032x210^2/2
=70.56J
divided that by heat of fusion of ice, 80cal/g
=0.882g
Its wrong tho
someone pls help :) thanks.[/B]
Are you talking about where i divided 70.56J by 80cal/g because i also tried converting 70.56J into cals and dividing it by 80cal/g again and it didnt work. If not please expand on what you mean buddySteamKing said:So you're just dividing numbers willy-nilly and disregarding units? Are calories the same as joules?
You didn't show any attempt at converting joules to calories or vice versa, so how are we to guess what you did? How do we know you even did the right conversion?Erenjaeger said:R u talking about where i divided 70.56J by 80cal/g because i also tried converting 70.56J into cals and dividing it by 80cal/g again and it didnt work. If not pls expand on wot u mean buddy
Presumably you are given this information for a reason. Eg don't just consider the bullets kinetic energy, include it's thermal energy too.Erenjaeger said:A lead bullet at 28.0°C
< Mentor Note -- text speak edited out of post >billy_joule said:Presumably you are given this information for a reason. Eg don't just consider the bullets kinetic energy, include it's thermal energy too.
Please, knock off the "text speak".Erenjaeger said:...
Can u pls help by explaining where I am going wrong.
thx
To recap:Erenjaeger said:so thermal energy is
Q=mcΔT
which works out to be 14.336 J
but how do i incorporate that into my working?
My working so far:
We= energy from bullet
=1/2 mv^2 + mc(sub)pΔT
mass in kg
the C(sub)p being specific heat of lead = 0.160 j/g degrees celsius
which comes out to be 70.57J
energy is fully absorbed by the blocked of ice
∴ We=Mi(Hf)
Mi = mass that melted
Hf = heat of fusion of ice
∴Mi=We/Hf
=211.97
which is wrong
Can u pls help by explaining where I am going wrong.
thx
We weren't given a specific heat of lead value, so I just googled itgneill said:I see that the value you've used for the specific heat of lead is 0.160 j/g/C, or 160 J/kg/C). Was this a given value in the problem or text that it came from? The reason I ask is that in the tables I've seen the value is generally given to be around 128 J/kg/C.
An example is the table of specific heats from the hyperphysics web site, where the entry for lead is 0.128 J/gm K. (gm = gram).Erenjaeger said:We weren't given a specific heat of lead value, so I just googled it
The impact of a lead bullet creates a sudden and intense release of energy, which in turn causes a rapid increase in temperature. This increase in temperature causes the ice to melt.
Yes, any type of bullet can cause ice to melt upon impact. However, bullets made of materials with higher melting points, such as copper or steel, may not cause as much melting as lead bullets.
Yes, the size of the bullet does play a role in the amount of ice melted. A larger bullet will typically have more mass and therefore more energy upon impact, resulting in a larger area of melted ice.
The temperature of the ice, the speed and angle of the bullet, and the composition of the ice (e.g. salt water vs. fresh water) can all impact the amount of ice melted by a lead bullet.
While the melting of ice by a lead bullet may seem concerning, it is not a significant environmental issue. The amount of ice melted by a single bullet is usually minimal and does not pose a threat to the environment.