A2.9 kg lump of aluminum is heated to 92 c and then dropped into 11.0 kg of water at 5.2 c. assuming that the lump–water system is thermally isolated, what is the system's equilibrium temperature? assume the specific heats of water and aluminum are 4186 and 900

9.85 C

Explanation:

Let T be the temperature off equilibrium mixture.

 Heat loss or gain = mass x specific heat x change in temperature

Heat loss by lump of Aluminium = 2.9 x 900 x (92 - T )

Heat gain by water = 11 x 4186 x ( T-5.2 )

Heat loss = heat gain

2.9 x 900 x (92 - T ) = 11 x 4186 x (T - 5.2 )

T = 9.85 C

answer: he got 23

explanation:

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.