How much ice is required to cool a 15.0oz drink from 77°F to 33°F, if the heat capacity of the drink is 4.18 J/g°C?

Question

When ice melts, it absorbs 0.33 kJ per gram. (Assume that the heat transfer is 100 % efficient.) Express answer using two significant figures.

Answer 1

I always say, "Start with what you've got, and go to what you want."

Here we are starting with 15.0 oz drink, and going to grams of ice to cool it down. Realize that we will have to convert ounces to grams, and °F to °C, along the way.

(15.0 oz drink)(28.35 g drink/oz drink)(40°F)(1°C/1.8°F)(4.18 J/g drink °C)(1 g ice/330 J)

Realize that all the units cancel except grams of ice. So when you multiply it out, you know that you'll have the correct answer.

115,060 grams of ice. That's 120 kilograms, or 4,100 ounces, using 2 significant digits.