A 10-kg iron ball is dropped onto a pavement from a height of 100 m. Suppose half of the heat generated goes into warming the ball. Show that the temperature increase of the ball is 1.1oC. (In SI units, the specific heat capacity of iron is 450 kJ/kg*oC.) Why is the answer the same for an iron ball of any mass?
Thanks for your help.
# Physics - MathMate, Thursday, September 24, 2009 at 6:53am
The basic principle is the conservation of energy.
When a ball is brought higher up by h=100 m., energy is required to make this happen. The energy is stored as potential energy, Ep=mgh, m=mass, g=acceleration due to gravity.
When the ball falls from this height in a free fall, energy is transformed into kinetic (movement) energy, given by the equation Ek=(1/2)mv², where v=velocity of the ball.
When the ball hits the ground, the velocity is reduced, so is the kinetic energy according to the formula Ek. Since total energy is conserved, the difference goes into other forms, such as kinetic energy of the fragments of soil flying around (sum of (1/2)mv² of the soil fragments), and the rest (in this case half) goes into heat, given by
Eh=mHΔT, H=specific heat, m again the mass, and ΔT=change in temperature.
I recapitulate:
The basic principle is the conservation of energy, in different forms.
Hope this clears up a little more of the picture. Post any time for more explanations.
# Physics - Ceres, Thursday, September 24, 2009 at 12:10pm
I worked on this the rest of last night and some more this morning and this is what I got.
Ep = mgh
Ep = 10kg(9.8m/s2)(100m)
Ep = 9800kgm2
I then equated that to the kenetic energy and used that amount to find the velocity:
9,800kg(m2) = 1/2 mv2
square root of 1960 = v
Now that I'm thinking it through more, I'm not sure why I did the calculations above.
Next I plugged in the numbers to find the change in temperature. I got:
Eh = mH(chng in Temp)
Eh = 10kg(450kj/kg*oC)(chng in T)
But I do not know the Eh (I'm not sure what that stands for) nor am I supposed to use the 1.1oC for the change in temp.
I understand all of the concepts you talked about - conservation of energy, potential energy to kenetic energy, etc. But my brain is still missing something and I'm not sure what it is. I included the above equations because maybe they would help you to see where my thinking has gone wrong. Thanks again for your help.
I'm sorry to hear that you're having trouble with the calculations. It seems like you're on the right track, but there are a few missing steps in your calculations.
To find the velocity of the ball when it hits the ground, you're correct in using the equation Ek = (1/2)mv^2. However, you forgot to include the potential energy in your calculation. Since the total energy is conserved, the kinetic energy is equal to the sum of the initial potential energy and the final kinetic energy. So the equation should be:
Ep = Ek
mgh = (1/2)mv^2
Canceling out the mass and rearranging the equation, you get:
v = sqrt(2gh)
Substituting the values of g (acceleration due to gravity) and h (height), you can find the velocity.
As for the change in temperature, you can use the equation Eh = mHΔT, where Eh represents the heat energy absorbed by the ball, m is the mass of the ball, H is the specific heat capacity of iron, and ΔT is the change in temperature.
Since half of the heat generated goes into warming the ball, you can use the equation:
Eh = (1/2)Ep
Substituting the value of Ep (calculated earlier) into the equation, you can find the heat energy absorbed by the ball.
Finally, rearranging the equation Eh = mHΔT, you can solve for ΔT, the change in temperature.
Now, the reason why the answer is the same for an iron ball of any mass is because the change in temperature is directly proportional to the mass of the ball. In other words, if you were to double the mass of the ball, the change in temperature would also double. This is because the heat energy absorbed is directly proportional to the mass of the ball.
To find the change in temperature of the iron ball, we need to consider the conservation of energy.
First, let's calculate the potential energy of the ball when it is raised to a height of 100m:
Ep = mgh
Ep = 10kg * 9.8m/s^2 * 100m
Ep = 9800kg m^2/s^2
When the ball falls, this potential energy is converted into kinetic energy. The formula for kinetic energy is:
Ek = 1/2 * mv^2
Since the ball is dropped from rest, the initial velocity is 0. Therefore, the initial kinetic energy is 0.
The total energy is conserved, so the potential energy is equal to the final kinetic energy when the ball hits the ground. Half of the heat generated goes into warming the ball, so the remaining energy is converted into heat.
Therefore, we have:
Ep = Ek + Eh
9800kg m^2/s^2 = 0 + Eh
Eh = 9800kg m^2/s^2
Now, we can calculate the change in temperature using the formula:
Eh = m * H * ΔT
Rearranging the formula, we can solve for ΔT:
ΔT = Eh / (m * H)
Plugging in the given values, we have:
ΔT = (9800kg m^2/s^2) / (10kg * 450 kJ/kg*°C)
Note: We need to convert the units of energy from J to kJ by dividing by 1000.
ΔT = (9800kg m^2/s^2) / (10kg * 450 kJ/kg*°C * 1000 J/kJ)
Simplifying, we get:
ΔT = 2.1778 °C
Therefore, the temperature increase of the iron ball is approximately 2.18 °C.
The answer is the same for an iron ball of any mass because the change in temperature is only dependent on the amount of energy transferred as heat and the specific heat capacity of the material. The specific heat capacity is a property of the material and is independent of mass. So, regardless of the mass of the iron ball, the same amount of energy is transferred as heat, resulting in the same change in temperature.
To find the change in temperature, we need to consider the energy conservation principle. The potential energy of the ball at the height of 100 m is equal to the kinetic energy just before it hits the ground. The difference in energy is then converted into heat.
Let's start by calculating the potential energy (Ep) of the ball:
Ep = mgh
where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (100 m).
Ep = 10 kg * 9.8 m/s^2 * 100 m
Ep = 9800 kg m^2/s^2
We can equate this potential energy to the kinetic energy just before the ball hits the ground. The formula for kinetic energy (Ek) is given by:
Ek = (1/2)mv^2
where v is the velocity of the ball just before hitting the ground.
So, Ek = (1/2) * 10 kg * v^2
Since energy is conserved, we can equate the potential energy (Ep) to the kinetic energy (Ek):
Ep = Ek
9800 kg m^2/s^2 = (1/2) * 10 kg * v^2
v^2 = 1960 m^2/s^2
v = sqrt(1960) m/s (taking the square root of both sides)
Now, we need to find the heat energy (Eh) generated when the ball hits the ground. Given that half of the heat generated goes into warming the ball, we can write:
Eh = (1/2) * m * H * ΔT
where H is the specific heat capacity of iron (450 kJ/kg*°C), m is the mass of the ball (10 kg), and ΔT is the change in temperature.
We need to solve for ΔT, the change in temperature. Rearranging the equation, we have:
ΔT = (2 * Eh) / (m * H)
Substituting the values we know, we get:
ΔT = (2 * 9800 kg m^2/s^2) / (10 kg * 450 kJ/kg*°C)
To convert kJ to J, we multiply by 1000:
ΔT = (2 * 9800 kg m^2/s^2) / (10 kg * 450 kJ/kg*°C * 1000 J/kJ)
ΔT = (19600 kg m^2/s^2) / (10 kg * 450000 J/°C)
ΔT ≈ 0.04355°C
Therefore, the temperature increase of the iron ball is approximately 0.04355°C. Based on the calculation, it appears that the answer would be the same for an iron ball of any mass because the mass cancels out in the equation.