Question

A calorimeter contains 195 g of water at 20.4 º C. A 37.8 g sample of an unknown metal is heated to 133 º C and placed into the water in the calorimeter. The final temperature of the metal and the water are 24.6 º C. What is the specific heat of the metal?

Answer 1

heat loss by the sample=heat gain by the water

Q=mass(kg)*Specific heat* Change in temperature

  =m*c*Change in T

0.0378*c*(24.6-133)*(-1)=0.195*4186*(24.6-20.4)

c=(0.195*4186*4.2)/(0.0378*(133-24.6))

 =836.7 J/(kg*Celsius degree)