From a height of 35.0 m, a 1.25 kg bird dives (from rest) into a small fish tank containing 50.0 kg of water.
What is the maximum rise in temperature of the water if the bird gives it all of its mechanical energy?
(delta)T= ?? degrees C
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To find the maximum rise in temperature of the water, we need to calculate the change in thermal energy of the water.
First, let's determine the initial potential energy of the bird. The potential energy formula is given by:
PE = m * g * h,
where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
Plugging in the given values:
m = 1.25 kg,
g = 9.8 m/s^2,
and h = 35.0 m.
PE = (1.25 kg) * (9.8 m/s^2) * (35.0 m)
= 431.25 J.
The potential energy of the bird is now converted into the thermal energy of the water. We can calculate the change in thermal energy using the specific heat formula:
q = m * c * ΔT,
where q is the change in thermal energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
m = 50.0 kg,
c = 4,186 J/kg·°C, (specific heat capacity of water)
Rearranging the formula, we get:
ΔT = q / (m * c).
Since the bird gives all its mechanical energy to the water, q is equal to the potential energy of the bird.
So, q = 431.25 J.
Now, we can plug in the values to find ΔT:
ΔT = (431.25 J) / ((50.0 kg) * (4,186 J/kg·°C)).
Calculating ΔT:
ΔT ≈ 0.0026 °C.
Therefore, the maximum rise in temperature of the water is approximately 0.0026 degrees Celsius.