Question

To calculate the specific heat capacity of some liquid, we have poured 235 g of water at a

temperature of 18.0°C into 250 g of this liquid at a temperature of 27.0 °C in a calorimeter with a
heat capacity of 210 J/K. The temperature stabilized at 20.5 °C. Using the specific heat capacity
of water, determine the specific heat capacity of the liquid. Assume that the liquid is not soluble
in water.

Answer 1

Solve for the specific heat capacity of the liquid using this equation:

Total heat = Cp water * mass water * ΔT1 + mass liquid * Cp liquid * ΔT2

Heat capacity * ΔTcalorimeter = Cp water * mass water * ΔT1 + mass liquid * Cp liquid * ΔT2

210 J/K (27-20.5) = (235 g * (18-20.5) C * 4.18 J/gK) + (250 * Cpliq * (27 -20.5) C )

solve for Cpliquid = 2.35 J/gK