Question

An ice cube of mass 9.0g is added to a cup of coffee, whose temperature is 90.0 °C and which contains 120.0 g of liquid. Assume the specific heat capacity of the coffee is the same as that of water. The heat of fusion of ice (the heat associated with ice melting) is 6.0kJ/mol. Find the temperature of the coffee after the ice melts. Use the equation given in the lab manual on page 86. Ignore the + (CC) (AT) since the question does not give that information and assumes no heat loss to the calorimeter. To be successful at this problem you must be very mindful of the signs and units. Leaving either out will cause you to get an incorrect answer. For the right side of the equation you cannot add KJ to J. Therefore be sure to convert 6.0 KJ/mol to 6.0 * 10 J/mol. Since delta H fusion is not in J/mol you must convet the grams of ice into moles of ice. A separate strategy could have been to convert the 6.0 KJ/mol to J/g and then that could have been multiplied by the mass of ice without having to modify the right side of ta equation. Here should be your final equation: -[(Tf-Ti) (s.h. H20/coffee) (mass H2O/coffee)] [ (AH fusion in J/mol) (mol ice)+ ((s.h. H2O) (mass ice) (Tf-0 °C)]

Answer 1

mass of ice = 9.0 g

moles of ice = (mass of ice) / (molar mass H₂O)

moles of ice = (9.0 g) / (18.0 g/mol)

moles of ice = 0.50 mol

heat gained by ice = (enthalpy of fusion) * (moles of ice)

heat gained by ice = (6.0 x 10³ J/mol) * (0.50 mol)

heat gained by ice = 3000 J

Let the final temperature of coffee = T_f

The equation used is

(mass coffee) * (specific heat coffee) * (T_i - T_f) = (heat gained by ice) + (mass ice) * (specific heat water) * (T_f - 0^oC)

Substituting the values,

(120.0 g) * (4.184 J/g.^oC) * (90.0 ^oC - T_f) = (3000 J) + (9.0 g) * (4.184 J/g.^oC) * (T_f)

45187.2 J - (502.08 J/^oC) * (T_f) = 3000 J + (37.656 J/^oC) * (T_f)

45187.2 J - 3000 J = (502.08 J/^oC) * (T_f) + (37.656 J/^oC) * (T_f)

42187.2 J = (502.08 J/^oC + 37.656 J/^oC) * (T_f)

42187.2 J = (539.736 J/^oC) * (T_f)

T_f = (42187.2 J) / (539.736 J/^oC)

T_f = 78.2 ^oC

The final temperature of coffee is 78.2 ^oC