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?
To determine the maximum temperature rise of the water balloon, we need to consider the change in the potential energy of the balloon as it falls. This change in potential energy will be converted into thermal energy, resulting in a temperature increase.
The potential energy of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
Given that the water balloon falls a distance of 21m, we can calculate the change in potential energy. However, to determine the maximum temperature rise, we need to consider the efficiency of energy conversion from potential energy to thermal energy.
Let's assume that the efficiency of energy conversion is 100%. In reality, some energy will be lost to air resistance and gravitational potential energy will not be fully transformed into thermal energy. However, for simplicity, we'll assume complete energy conversion.
The change in potential energy is given by ΔPE = mgh, where ΔPE is the change in potential energy, m is the mass of the water balloon, g is the acceleration due to gravity, and h is the height.
Since we want to find the maximum temperature rise, we can assume that all the potential energy is converted into thermal energy. In other words, ΔPE = ΔThermal Energy.
The formula for thermal energy change (ΔQ) is ΔQ = mCΔT, where ΔQ is the thermal energy change, m is the mass, C is the specific heat capacity of water, and ΔT is the temperature change.
Since ΔPE = ΔThermal Energy and ΔThermal Energy = ΔQ, we can equate the two expressions:
mgh = mCΔT
Simplifying the equation, we can cancel out the mass (m) term:
gh = CΔT
Rearranging the equation, we can solve for the temperature change (ΔT):
ΔT = (gh) / C
Now, we can substitute the known values into the equation:
ΔT = (9.8 m/s²)(21m) / (specific heat capacity of water)
The specific heat capacity of water is approximately 4.18 J/g°C.
Converting the units:
ΔT ≈ (9.8 m/s²)(21m) / (4.18 J/g°C)
Calculating the result:
ΔT ≈ 49.0476 °C
Therefore, the maximum temperature rise of the water balloon due to being dropped is approximately 49.05°C.