A 2.9 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