(a)The angular speed of a ceiling fan is increased uniformly in 3.0 s from 180 revs/min to 290 revs/min. What is the angular displacement of the blades during the acceleration?
(b) 25 g of copper at 430 K is mixed with 150 mL of water 290 K. Assuming energy is exchanged only between the copper and water, what is the final temperature of the mixture? The specific heats of copper and water are respectively 386 /kg.k and 4186 /kg.k
We will gladly critique you work, when shown.
(a) To find the angular displacement of the blades during the acceleration, we first need to calculate the initial and final angular speeds in radians per second.
Given:
Initial angular speed, ω1 = 180 revs/min
Final angular speed, ω2 = 290 revs/min
To convert the revolutions per minute (revs/min) to radians per second (rad/s), we use the following conversion factor:
1 rev = 2π radians
1 minute = 60 seconds
So, we can convert the initial and final angular speeds as follows:
Initial angular speed, ω1 = (180 revs/min) * (2π rad/rev) / (60 s/min)
Final angular speed, ω2 = (290 revs/min) * (2π rad/rev) / (60 s/min)
Next, we can calculate the average angular speed during the acceleration:
Average angular speed, ω_avg = (ω1 + ω2) / 2
Now, we can calculate the angular displacement using the formula:
Angular displacement, θ = ω_avg * t
Given:
Time, t = 3.0 s
Substituting the values, we have:
Angular displacement, θ = ω_avg * t = (ω1 + ω2) / 2 * t
Now, we can calculate the angular displacement by substituting the values and evaluating the equation.
(b) To find the final temperature of the mixture, we need to use the principle of energy conservation. The heat lost by the copper is equal to the heat gained by the water.
First, let's find the heat lost by the copper using the formula:
Qcopper = mcopper * ccopper * ΔTcopper
where:
mcopper = mass of copper = 25 g
ccopper = specific heat of copper = 386 J/kg·K
ΔTcopper = change in temperature of copper = Tf - Tc
Now, let's find the heat gained by the water using the formula:
Qwater = mwater * cwater * ΔTwater
where:
mwater = mass of water = 150 g = 150 mL
cwater = specific heat of water = 4186 J/kg·K
ΔTwater = change in temperature of water = Tf - Tw
Since the energy lost by copper is equal to the energy gained by water, we can set the two equations equal to each other:
Qcopper = Qwater
Simplifying, we have:
mcopper * ccopper * ΔTcopper = mwater * cwater * ΔTwater
Now, we can substitute the values and solve the equation for Tf (final temperature).