When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 12.0-oz drink from \(75\ ^{\circ}\mathrm{F}\) to \(35\ ^{\circ}\mathrm{F}\), if the heat capacity of the drink is \(4.18\mathrm{\ J}/\mathrm{g}^{\circ}\mathrm{C}\)? (Assume that the heat transfer is 100% efficient.)

Equation Transcription:

Text Transcription:

75 deg F

35 deg F

4.18 J/g deg C

Solution 106P

In this question it has given that the heat capacity of the drink= 4.18 J/g °C

Heat absorbs (q) = 0.33 KJ per gram

T2= 75oF

T1 = 35oF

= T2 -T1 = (75-35)oF = 40oF

It is known that,

1.0 oz = 28.3g.

Hence 12.0 oz will be equal to,

12.0 oz x 28.35g/oz = 339.6g (m)

The heat absorbed (q) = mCΔT

= (339.6 g) x (4.184 J/g C) x (40 )

= 56.8 kJ 1 g 56.8 kJ x

= 170 g of ice 0.33 kJ