Question

A sample of aluminum, which has a specific heat capacity of

0.897·J·g−1°C−1

, is put into a calorimeter (see sketch at right) that contains

300.0g

of water. The aluminum sample starts off at

94.5°C

and the temperature of the water starts off at

21.0°C

.When the temperature of the water stops changing it's

23.8°C

. The pressure remains constant at

1atm

.

Calculate the mass of the aluminum sample. Be sure your answer is rounded to

2

significant digits.

Answer 1

Mass of water, m = 300 g Specificheat of water, s = 4.184 J/g. 'C Change in temperature AT = 23.8°C-21.0°C = 2.8°C Amount of heat absorbed by water, water = mSAT = 3008x4.184EX(2.8°c) = 3514.56 J Heat released by aluminum is equal to heat gained by water We known that, beim =- mer = -3514.56 J Change in temperature, AT = 23.8°C-94.5°C = -70.7°C Specificheat of aluminum ,s = 0.897 J/g °C mass of aluminum, m = Palmtom SAT -3514.56 J (0.897 )(-7070) = 55.4 g s 55 g