Question

From a branch 35 m high, a 0.75 kg bird dives into a small fish tank containing

Answer 1

Answer:

ΔT = 1.22*10^-3 °C

Explanation:

First, you calculate the potential energy of the bird when it is at 35 m high. The potential energy is also the mechanical energy of the bird in this case.

$U=mgh$

m: mass of the bird = 0.75kg

g: gravitational constant = 9.8m/s^2

h: height = 35m

$U=(0.75kg)(9.8m/s^2)(35m)=257.25\ J$

All this energy is given to the water. You use the following formula in order to calculate the change in temperature:

$Q=mc\Delta T$

m: mass of the water = 50kg

c: specific heat of water = 4186 J/kg°C

Q is equal to U (potential energy of the bird) because the bird gives all its energy to water. By doing ΔT the subject of the formula you obtain:

$\Delta T=\frac{Q}{mc}=\frac{257.25J}{(50kg)(4186J/kg°C)}=1.22*10^{-3}\ \°C$

hence, the maximum rise in temperature is 0.00122 °C