Question

A calorimeter is filled with 28.3g of radiator coolant, ethlylene glycol at 20.1 degrees Celsius. A 15.2 g block of Al at 86.2 degrees celsius is placed into the calorimeter. Assuming the heat capacity of the calorimeter is neglible what will be the final temperature for the coolant and the aluminum block. (c of ethylene glycol -2.42 J/mol g*k, c of aluminum = .900J/g*K)

Answer 1

The heat given up by the block of aluminum = the heat gained by the coolant.
Let x by the final temperature.
Heat given up by the aluminum block = 0.0152 kg * 0.900 J/gC * (86.2C - X) = 0.0283 kg * 2.42 J/gC * (X - 20.1C)

Solving for X,

X=31.10 C

The final temperature of Al block will be 31.10 C