How Much Ice Melts When a Hot Bullet Hits It?

[PrudensOptimus]

A 1.0 g lead bullet at 33°C is fired at a speed of 250 m/s into a large block of ice at 0°C, in which it becomes embedded. What quantity of ice melts?


Heres what I have:

From ΔQ = mcΔT,

m_A = dQ/[c(dT)]

And from ΔQ_A = -ΔQ_B

...

Further cogitations are in process. Meanwhile please respond, thanks.

 

[KLscilevothma]

1. The kinetic energy in the bullet is changed to internal energy (increase in temperature) when it stops inside the ice block. You can use the following formula to calculate the change in temperature of the bullet.
1/2 mv² = mc_LΔT
So the final temperatre of the bullet when embeded inside the ice block = 33 + ΔT

2. Assume the temperature in the ice block stays at 0°C.
The final temperature of the bullet will be 0°C, which means the energy released when the bullet is cooled from (33 + ΔT)°C to 0°C is used to melt the ice block.
So you can use the following formula to find out the quantity of ice melted.
m_Lc_L(33 + ΔT) = m_iceL_ice

Where
m_L = mass of the lead bullet
c_L = specific heat capacity of the bullet
m_ice = mass of ice melted
L_ice = latent heat of fusion of ice

 

[PrudensOptimus]

Originally posted by KLscilevothma
1. The kinetic energy in the bullet is changed to internal energy (increase in temperature) when it stops inside the ice block. You can use the following formula to calculate the change in temperature of the bullet.
1/2 mv² = mc_LΔT
So the final temperatre of the bullet when embeded inside the ice block = 33 + ΔT

2. Assume the temperature in the ice block stays at 0°C.
The final temperature of the bullet will be 0°C, which means the energy released when the bullet is cooled from (33 + ΔT)°C to 0°C is used to melt the ice block.
So you can use the following formula to find out the quantity of ice melted.
m_Lc_L(33 + ΔT) = m_iceL_ice

Where
m_L = mass of the lead bullet
c_L = specific heat capacity of the bullet
m_ice = mass of ice melted
L_ice = latent heat of fusion of ice

How is the Final Temperature of the Lead bullet 33 + ΔT?

Perhaps you made a typo?

By solving the KE = Q equation, i found T_f = (v^2/2c) + 33Celsius.

 

[PrudensOptimus]

93.781437125748502994011976047904 grams is the answer i got. not sure if i did it right or not.

 

[PrudensOptimus]

ok the answer is 0.107g...

can someone do this problem with another approach pls?

 

[ShawnD]

Originally posted by PrudensOptimus
A 1.0 g lead bullet at 33°C is fired at a speed of 250 m/s into a large block of ice at 0°C, in which it becomes embedded. What quantity of ice melts?

All the changes in energy add up to 0 so base an equation around that using all changes in energy.

$$mc \Delta T + mL_v - \frac{1}{2}mv^2 = 0$$

Check your signs. Terms gaining energy should be positive, losing energy should be negative. The velocity part is a loss in energy so that term is negative. Changes in temperature will always work themselves out with the proper sign; just be careful of latent heats and kinetic energy.

mcT is the bullet's temperature change, mLv is the ice melting and 1/2mv^2 is the kinetic energy of the bullet. From there just start filling in the equation. The only variable is the mass of ice melting.