Question

A 485.4-g sample of an element at 192°C is dropped into an ice–water mixture; 117.0 g of ice melts and an ice–water mixture remains. Calculate the specific heat of the element. ΔHfusion = 6.02 kJ/mol (for liquid water at 0°C). Specific heat =______ J/g∙°C

Answer 1

here the ice-water mixture remains at the end, so the final temperature of system = 0°C

Let the specific heat of the element be Cp

Heat released by element = 485.4 x Cp x (192 - 0)

                                         = 93196.8 Cp J

Moles of ice melted = 117.0 / 18.02 = 6.493 mol

Heat absorbed by ice melting = 6.493 x 6.02 = 39086.57 J

Heat released = heat absorbed

93196.8 Cp = 39086.57

Specific heat = 0.419 J/g°C