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

15.3 cos 65 degrees trig

grav potential energy