Question

The specific heat capacity of an unknown metal is to be found using calorimetry. A 35.0g sample of the metal is placed into a boiling water bath at a temperature of 100.0ºC. In the calorimeter lies 101.1g of water at a temperature of 22.1ºC. After thorough heating of the metal, it is transferred to the calorimeter. The final temperature of the metal/water system is 22.9ºC. Calculate the specific heat capacity of the metal.

Answer 1

mass of sample = 35.0 g

iinitial temperature = 100 C

principle of calorimeter is

heat gained= heat loss

now final temp after putting the metal into the water = 22.9 C

now heat gained by the water

m=mass of water , c = specific heat capacity of water = 1 Calorie g-1 c-1 change in temperature = 22.9-22.1=0.8

=m*c*change in temperature

=101.1*1*0.8=80.88 calories

now at first heat lost by metal

maas of metal = 35.0 g , c2=specific heat of metal , change in temperature = here initially the temperature of the metal is 100 because it is thoroughly heated (100-22.9)=77.1C

now m*c*change in temperature

35.0* c2*77.1=2698.5 c2

it is equal to 80.88 calories

2698.5 c2 = 80.88 cal

c2 = 80.88/2698.5=0.02997 Calorie g-1 C-1