Question

A copper calorimeter of heat capacity 32 J/°C contains 100 g of oil at 30°C. 80 g of a metal of sp. heat cap. 0.12 J/g°C at 90°C is dropped into the oil and final temperature is 35°C. Calculate the sp. heat cap. of the oil.

Answer 1

Heat lost by metal = Heat gained by calorimeter and oil

m₃C₃(x-z) = m₁C₁(z-y) + m_zC_z(z-y)

where, m₁C₁ = 32 J/°C
m₂ = 100 g
y = 30°c
m₃ = 80 g
C₃ = 0.12 J/g°C
x = 90°C
z = 35°C
⇒ 80 x 0.12 x (90 - 35) = 32x (35 - 30) + l00 x C₂ x (35 - 30)
⇒ 528 = 160 + 500C₂
⇒ C₂ = 0.736 J/ g°c