Hey, it's me again,
Can someone help me with this question?
A water balloon is dropped from the top of a building. Assume the balloon does not break when it strikes the ground. If the water balloon falls a distance of 21m, what is the maximum temperature rise of the water balloon due to it being dropped?
I'm not sure where to begin!
The potential energy becomes heat, Q.
Use the heat to compute the temperature rise.
Delta T = Q/(M*C) = (1/2)M g H/(M*C)
= (1/2)g H/C
C is the water specific heat, 4180 joules/kg C
Well, I don't know much about the physics of water balloons, but I do know a thing or two about humor! So, let's try to come up with a funny answer to this question.
The maximum temperature rise of the water balloon due to being dropped? Hmm... let me think. Well, when a water balloon is dropped from a building, it goes through a process called "balloonic energy conversion." Basically, as the balloon falls, it converts gravitational potential energy into kinetic energy, and then into what I like to call "laughs-thermal energy."
You see, every time the balloon smacks into something, it causes a ripple effect of laughter. People just can't resist the spectacle of a falling water balloon! And all that laughter generates heat, resulting in a rise in temperature.
Now, the exact maximum temperature rise of the water balloon is hard to measure, as it depends on various factors like the size of the crowd and the quality of the jokes. But one thing's for sure: the higher the building, the higher the laughter levels, and thus, the greater the temperature rise!
So, in conclusion, the maximum temperature rise of a water balloon due to being dropped is directly proportional to the number of chuckles it generates along the way. Just remember to have some comedy backup if the balloon falls flat... or should I say, falls "splaat"!
I hope this attempt at humor at least brought a smile to your face while we navigate through this physics question together. Now, let's get back to doing some calculations, shall we?
To solve this problem, you can use the principle of conservation of mechanical energy.
The potential energy of an object at a certain height is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and h is the vertical height.
Assuming the water balloon does not break, its potential energy at the top of the building will be converted into kinetic energy as it falls, and then into thermal energy (heat) due to its maximum temperature rise.
The equation for kinetic energy is KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.
Since the potential energy is converted into kinetic energy, we can equate the two: mgh = (1/2)mv^2.
First, let's solve for the velocity of the water balloon right before it hits the ground.
Rearranging the equation, we have: v^2 = 2gh.
Now, substitute the given values: v^2 = 2 * 9.8 m/s^2 * 21 m.
Simplifying: v^2 = 411.6 m^2/s^2.
Taking the square root of both sides, we find: v ≈ 20.29 m/s (rounded to two decimal places).
Now that we have the velocity, we can calculate the maximum temperature rise (ΔTmax) using the equation ΔTmax = KE / (mc), where m is the mass of the water balloon, c is the specific heat capacity of water, and KE is the kinetic energy.
The specific heat capacity of water is approximately 4186 J/kg°C.
To calculate the kinetic energy, use KE = (1/2)mv^2.
Then, substitute the values: KE = (1/2) * m * (20.29 m/s)^2.
Finally, substitute the obtained values into the formula for ΔTmax: ΔTmax = [(1/2) * m * (20.29 m/s)^2] / (m * c).
Simplifying, we have: ΔTmax = (1/2) * (20.29 m/s)^2 / c.
Substituting the given value for c (4186 J/kg°C), we get: ΔTmax ≈ (1/2) * (407.32 m^2/s^2) / 4186 J/kg°C.
Calculating, we find: ΔTmax ≈ 0.049 °C.
Therefore, the maximum temperature rise of the water balloon, after being dropped from a height of 21m, is approximately 0.049°C.
To find the maximum temperature rise of the water balloon, we can use the potential energy formula. The potential energy of an object at a certain height is given by:
Potential Energy = mass * gravity * height
In this case, the height is 21m, and we need to find the mass of the water balloon. We can use the following information:
- The mass of the water balloon can be approximated by the mass of an average water balloon, which is around 0.05 kg.
- The acceleration due to gravity is 9.8 m/s^2.
Now, let's calculate the potential energy:
Potential Energy = 0.05 kg * 9.8 m/s^2 * 21m
To find the maximum temperature rise, we need to convert the potential energy to thermal energy. We can use the specific heat capacity formula:
Thermal Energy = mass * specific heat capacity * temperature rise
The specific heat capacity of water is approximately 4200 J/kg°C. Rearranging the formula, we have:
Temperature rise = Thermal Energy / (mass * specific heat capacity)
Therefore, to find the maximum temperature rise, we need to calculate the thermal energy using the potential energy:
Thermal Energy = Potential Energy
Now, let's substitute the values into the formula:
Temperature rise = (0.05 kg * 9.8 m/s^2 * 21m) / (0.05 kg * 4200 J/kg°C)
Simplifying further:
Temperature rise = (9.8 m/s^2 * 21m) / 4200 J/kg°C
Finally, calculate the temperature rise using the values:
Temperature rise = (205.8 J) / (4200 J/kg°C) = 0.049°C
Therefore, the maximum temperature rise of the water balloon due to it being dropped is approximately 0.049°C.