A 20 g lead bullet leaves a rifle at a temperature of 47 degrees celcieus and travels at a velocity of 500 m/s until it hits a large block of ice at 0 degrees celcuis and comes to rest within it. How much ice will melt?
KEbullet+heatlosstocoolbullet=massice*Hf
1/2 m v^2+ m(specheatlead)(47)=Massice*Heatfusionice
To calculate the amount of ice that will melt, we need to consider the heat transferred from the lead bullet to the ice block until it reaches the same temperature.
First, let's find the initial heat energy of the bullet using the formula:
Q = mcΔT
Where:
Q is the heat energy
m is the mass of the bullet
c is the specific heat capacity of lead
ΔT is the change in temperature
Given:
Mass of the bullet (m) = 20 g = 0.02 kg
Initial temperature of the bullet (T1) = 47°C = 320 K
Final temperature of the bullet (T2) = 0°C = 273 K
Specific heat capacity of lead (c) = 128 J/kg°C
Using the formula, we can calculate the initial heat energy (Q):
Q = mcΔT
Q = (0.02 kg) * (128 J/kg°C) * (273 K - 320 K)
Q = (0.02 kg) * (128 J/kg°C) * (-47 K)
Q = -120.32 J
The negative sign indicates that energy is lost from the bullet and gained by the ice block. The bullet transfers this energy to the ice as heat.
The heat energy transferred to the ice (Q) is then used to melt a certain amount of ice, which can be calculated using the latent heat of fusion.
The latent heat of fusion (L) for ice is the amount of energy required to convert one gram of ice at 0°C into water at 0°C. Its value is 334 J/g.
Next, let's calculate the amount of ice melted (melted mass) using the formula:
melted mass = Q / L
Given:
Q = -120.32 J
L = 334 J/g
melted mass = (-120.32 J) / (334 J/g)
Converting J to grams (1 g = 1 J):
melted mass = (-120.32 g) / (334 g)
melted mass ≈ 0.36 g or 0.00036 kg
Therefore, approximately 0.36 grams (or 0.00036 kg) of ice will melt when the lead bullet hits the ice block and comes to rest within it.