HELP ! ! ! Chemistry home work . ! ! ! ! ! ! !

Question

A hot lump of 37.4g of aluminum at an initial temperature of 63.3C is placed in 50ml of h2O initially at 25C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water given that the specific heat of aluminum is .903j/(g•C)? Assume no heat is lost to surroundings

Answer 1

heat lost by aluminum = heat gained by water

At thermal equilibrium, temperature of aluminum = temperature of water

q = mC∆T where q is heat; m is mass; C is specific heat and ∆T is the change in temperature.

Assuming the density of water is 1 g/ml, the mass of water will be 50 g. And using C_H2O = 4.184 J/g/deg...

(37.4 g)(0.903 J/g/deg)(63.3 - T_f) = (50 g)(4.184 J/g/deg)(T_f - 25)

33.77(63.3 - T_f) = (209.2)(T_f-25)

2138 - 33.77T_f = 209.2T_f - 5225

242.97T_f = 73 63

T_f = 30.3ºC final temperature of both the aluminum and the water