Question

How much ice is required to cool a 13.0 oz drink from 71 ∘F to 32 ∘F, if the heat capacity of the drink is 4.18 J/g∘C? (Assume that the heat transfer is 100 % efficient.)

Answer 1

When ice melts, it absorbs 0.33 kJ energy for 1 gram.
1 oz = 28.3 g. It means 13.0 oz will be 367.9 g of drink.
Q = mCΔT.
ΔT = 26.67 oC.
Q = (367.9g)*(4.18 J/goC)*(26.67oC) = 41014 J energy = 41.014 kJ.
41.014=1/10^3 J * 1g/0.33kJ = 124 g of ice is required.