1. How much ice at -10 degrees Celsius is required to cool a mixture of 0.1 kg ethyl alcohol and 0.1 kg water from 20 degrees Celsius to 0 degrees Celsius?
2. Find the heat produced by 1 KW heating element in one hour.
3. A 3g lead bullet at 30 degrees Celsius is fired at a speed of 240 m/s into a large block of ice at 0 degrees Celsius in which it becomes embedded. What quantity of ice melt?
thanks. :)
1. Write an energy balance equation: Heat absorbed by heating (from -10) and melting X g of ice equals heat removed from the liquid water and alcohol as they is lowered to 0 C. Solve for the amount of melting ice, X
2. Watts is Joules per seond. Multiply 1000 W by the number of seconds
3. First compute the number of Joules of kinetic energy in the bullet. It will equal the heat available to melt, Q. Assume a final equilibrium of 0 C and write an equation saying that much Q lowers the temp of the lead to 0 C while melting X g of ice. X will be the only unknown in that equation. Solve for it.
thanks. :)
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.