Answer to Question #114193 in Physical Chemistry for Tom

Question

74.8 g of copper at 143.20C is immersed in 165 mL of glycerin (C3H8O3, dglycerin = 1.26g / mL) at 24.80C in an insulated container. The final temperature is measured at 31.10C. Accordingly, what is the heat capacity of glycerine in the unit of J. mol-1.0C-1? (c copper = 0.385 J. mol-1.0C-1)

Answer 1

Heat (Q) is a product of heat capacity (c), number of moles (n) and temperature change:

Q = c*m*(T₂-T₁)

The same heat was lost by copper and gained by glycerol, hence:

c(Cu)*n(Cu)*(143.20-31.10) = c(glycerol)*n(glycerol)*(31.10-24.80)

Here:

n(glycerol) = 165 ml * 1.26 g/ml / 92.09 g/mole = 2.258 mole

n(Cu) = 74.8 g / 63.546 g/mole = 1.177 mole

Find heat capacity of glycerol:

c(glycerol) = 0.385*1.177*(143.20-31.10)/(2.258*(31.10-24.80)) = 3.571 J/(mol*C)

Answer: 3.571 J/(mol*C)