Question

A student obtains the following data in a calorimetry experiment designed to measure the specific heat of aluminum.

Initial temperature of water and calorimeter 70.4°C

Mass of water 0.403 kg

Mass of calorimeter 0.04 kg

Specific heat of calorimeter 0.60 kJ/kg·°C

Initial temperature of aluminum 27.1°C

Mass of aluminum 0.196 kg

Final temperature of mixture 66.4°C

(a) Use these data to determine the specific heat of aluminum. J/kg · °C

(b) Is your result within 15% of 900 J/kg · °C? (Yes/ No)

Answer 1

We know that

Heat released by water = Heat gained by Aluminum

Q1 = Q2

Q1 = Qw + Qc

Qw = heat released by water = Mw*C*dT

Qc = Heat released by calorimeter = Mc*Cc*dT

dT = 70.4 - 66.4 = 4 C

C = Specific heat of water = 4186 J/kg-C

Cc = Specific heat of calorimeter = 600 J/kg-C

Mw = mass of water = 0.403 kg

Mc = mass of calorimeter = 0.04 kg

Q2 = Heat gained by aluminum = Ma*Ca*dT

dT = 66.4 - 27.1 = 39.3

Ma = mass of aluminum = 0.196 kg

So,

Q1 = Q2

0.403*4186*4 + 0.04*600*4 = 0.196*Ca*39.3

Ca = (0.403*4186*4 + 0.04*600*4)/(0.196*39.3)

Ca = 888.48 = Specific heat of aluminum

B.Yes result is within 15% of 900 J/kg-C