1. To solve this problem, we need to calculate the heat exchanged between the mixture and the ice. Here are the steps:
Step 1: Calculate the heat lost by the mixture of ethyl alcohol and water.
Q = mcΔT
Where:
Q = heat lost (in Joules)
m = mass of the mixture (in kg)
c = specific heat capacity of the mixture (in J/kg°C)
ΔT = change in temperature (in °C)
Given:
m = 0.1 kg
ΔT = (20 - 0) = 20°C
Specific heat capacity of the mixture (c) can be approximated as the average of ethyl alcohol and water:
c ≈ (c_ethanol + c_water) / 2
The specific heat capacity of ethyl alcohol (c_ethanol) is approximately 2.45 J/g°C, which is equivalent to 2,450 J/kg°C.
The specific heat capacity of water (c_water) is approximately 4.18 J/g°C, which is equivalent to 4,180 J/kg°C.
Using the approximation formula:
c ≈ (2,450 + 4,180) / 2 ≈ 3,315 J/kg°C
Plugging the values into the formula:
Q = 0.1 kg * 3,315 J/kg°C * 20°C = 6,630 J
Step 2: Calculate the heat gained by the ice.
The heat gained or absorbed by the ice is equal to the heat lost by the mixture because energy is conserved.
Therefore, the heat gained by the ice is also 6,630 J.
Step 3: Calculate the mass of ice required to absorb the heat gained.
Using the latent heat of fusion of ice (L_f), which is 334,000 J/kg:
Q = mL_f
Solving for m (mass):
m = Q / L_f = 6,630 J / 334,000 J/kg ≈ 0.0198 kg (or 19.8 g)
So, approximately 19.8 grams of ice at -10 degrees Celsius are required to cool the mixture from 20 degrees Celsius to 0 degrees Celsius.
2. To find the heat produced by a 1 kW heating element in one hour, we can use the formula:
Heat (in Joules) = Power (in Watts) × Time (in seconds)
Given:
Power (P) = 1 kW = 1,000 W
Time (t) = 1 hour = 3,600 seconds
Using the formula:
Heat = 1,000 W × 3,600 s = 3,600,000 J
Therefore, the heat produced by the 1 kW heating element in one hour is 3,600,000 Joules.
3. To find the quantity of ice melted when a bullet is fired into it, we need to calculate the heat transferred from the bullet to the ice.
Step 1: Calculate the heat lost by the bullet.
Q = mcΔT
Where:
Q = heat lost (in Joules)
m = mass of the bullet (in kg)
c = specific heat capacity of lead (in J/kg°C)
ΔT = change in temperature (in °C)
Given:
m = 3 g = 0.003 kg
ΔT = (30 - 0) = 30°C (assuming the bullet starts at the same temperature as the surrounding air)
The specific heat capacity of lead (c) is approximately 130 J/kg°C.
Plugging the values into the formula:
Q = 0.003 kg * 130 J/kg°C * 30°C = 11.7 J
Step 2: Calculate the heat gained by the ice.
The heat gained by the ice is equal to the heat lost by the bullet (assuming no heat is lost to the surroundings) because energy is conserved.
Therefore, the heat gained by the ice is also 11.7 J.
Step 3: Calculate the mass of ice melted by converting the heat gained into the heat required to melt the ice.
Using the latent heat of fusion of ice (L_f), which is 334,000 J/kg:
Q = mL_f
Solving for m (mass):
m = Q / L_f = 11.7 J / 334,000 J/kg ≈ 0.035 g (or 35 mg)
So, approximately 35 milligrams of ice would melt when the 3g lead bullet is fired into it at a speed of 240 m/s and becomes embedded in the ice.