A liquid of density 0.85 g/cm3 flows through a calorimeter at the rate of 8.0 cm3/s. Heat is added by means of a 250 W electric heating coil and a temperature difference of 15oC is established in steady-state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will be

This is a problem on 'flow calorimeter' used to measure specific heat of a liquid.Amount of heat supplied to the water per second by the heating coil =QS=250J=2504186kcalThe volume of liquid flowing out per second 1=8.0cm3                                                                        =8×10−6m3Mass of this liquid =(0.85)×1000×8×10−6kgTemperature rise of this mass of liquid =15∘CHence, 2504186=mst=0.85×8×10−3×s×15Hence, s=250×1034186×0.85×8×15=0.6kcal/kgK