Question

calculate the heat of fusion of ice from the following data at 0°c add to water :
Mass of calorimetet 60 g
Mass of calorimeter and water 460
Mass of calorimeter plus water and ice 618 g
Initial temperature of water 38 °c
Final temperature of mixture 5 °c
Specific heat of calorimeter 0.10 cal/g.°c

Answer 1

80 cal/g

Answer 2

first begin by identify the mass of every given substance.
mass calorimeter = 60g
mass water = 460g-60g= 400g
mass ice= 618g-460g=158g
Qgain=Qlost
Qice=Qwater
therefore
mL + m(detlta temp)C=(delta temp)(mass water * C +mass calor *C)

and then we make L subject of the formula
so
L=[33(0.4*4186 + 0.06*400)-(0.158*5*4186)] / 0.158
L=333799.1139 J/kg