Question

At a local convenience store, you purchase a cup of coffee, but, at 98.4°C, it is too hot to drink. You add 28.5 g of ice that is −2.2°C to the 248 mL of coffee. What is the final temperature of the coffee? (Assume the heat capacity and density of the coffee are the same as water and the coffee cup is well insulated.)

Answer 1

If a substance is not changing phase, use q = mCΔT
If a substance (like ice) does change phase, use q = nΔHfus

Ice (-2.2*C -----> Ice (0*C) no phase change

q =mCΔT
= (28.5 g)(2.09J/g*C)(0*C -(-2.2*C))
q = 131.043 J = 0.131043 kJ

Ice (0*C) ---------> Water (0*C) phase change

q = nΔHfus
= (1.58mol)(6.02kj/mol) [28.5 g Ice / 18.o g/mol = 1.58 mol H2O]
= 9.532 kJ

Total amount of heat gained by the ice = 9.532 kJ + 0.131043 kJ = 9.663043 kJ = 9663.043 J

Heat gained by ice = Heat lost by water

q = mC_wΔT

-9663.043 J= (248 g)(4.184 J/g*C)(T_f- 98.4 C)

T_f = 89.087 ^oC

Final temperature of the coffee = 89.1 ^oC

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