A 1500 kg car is travelling at 20 m/s. It brakes to a stop causing the brake pads to heat up from 25 degrees Celsius to a temperature too hot to touch. The total mass of the pads altogether is 4.0 kg and their heat capacity is 8.9 x 102
J/kg0C. If these pads are removed from the car while hot and put into a bucket containing 8.5 kg of water at 20 0C,what final temperature will the water reach? (Assume all the energy transfers are 100% efficient.) The heat capacity
of water is 4.2 x 103 J/kg0C.
E = ke = (1/2) m v^2
that goes into the brake pads and into the water heating both up to T
(1/2)1500 (400) = 4200*8.5 (T-20)+ 890*4(T-25)
To find the final temperature that the water will reach after the hot brake pads are added to it, you can apply the principle of conservation of energy.
First, calculate the energy transferred from the brake pads to the water. This can be done using the equation:
Q = mcΔT
Where Q is the energy transferred (in joules), m is the mass (in kg), c is the specific heat capacity (in J/kg°C), and ΔT is the change in temperature (in °C).
For the brake pads, the initial temperature is 25°C and the final temperature is the temperature too hot to touch. The mass of the brake pads is given as 4.0 kg, and their specific heat capacity is 8.9 x 10^2 J/kg°C. Therefore, the energy transferred from the brake pads is:
Q_brake_pads = 4.0 kg * 8.9 x 10^2 J/kg°C * (T_final - 25°C)
Next, calculate the energy absorbed by the water using the same equation. The initial temperature of the water is 20°C, and its mass is given as 8.5 kg. The specific heat capacity of water is 4.2 x 10^3 J/kg°C. Therefore, the energy absorbed by the water is:
Q_water = 8.5 kg * 4.2 x 10^3 J/kg°C * (T_final - 20°C)
Since the energy transferred from the brake pads is equal to the energy absorbed by the water (assuming 100% efficiency), we can set the two equations equal to each other:
4.0 kg * 8.9 x 10^2 J/kg°C * (T_final - 25°C) = 8.5 kg * 4.2 x 10^3 J/kg°C * (T_final - 20°C)
Now, you can solve this equation to find the final temperature (T_final) of the water. Simplifying the equation gives:
35.6 x (T_final - 25) = 35.7 x (T_final - 20)
35.6T_final - 891 = 35.7T_final - 714
0.1T_final = 177
T_final = 177 / 0.1
T_final ≈ 1770°C
Therefore, the final temperature that the water will reach after adding the hot brake pads is approximately 1770°C.