When ice melts, it absorbs 0.33 kJ per gram.

How much ice is required to cool a 13.0 oz drink from 71 ∘F to 39 ∘F, if the heat capacity of the drink is 4.18 J/g∘C? (Assume that the heat transfer is 100 % efficient.)

sum of heats gained is zero

13oz*28.3g/oz*4.18J/gC*deltatemp+33oJ/g*massice=0

now for delta temp, the change in F is 71-39=22F, and in C that is 22*5/9

solve for mass of ice.

I actually found the answer. The answer is 83 grams.

You must convert the values accordingly. oz to grams, Fahrenheit to Celsius.
You calculate the heat lost by the drink first. This is the setup:
=(368.544g)(4.18J/gC)(-17.7778C)
the change in temp is found by first converting your values to Celsius, then subtracting the initial from the final.
after you find the answer to the above equation, you divide it by the 0.33kJ that ice absorbs when it melts. You must convert the kJ to J first though.
this will give you the number of grams needed to cool the drink.

To calculate the amount of ice required to cool the drink, we need to determine the amount of heat energy transferred from the drink to the ice.

First, we need to convert the given temperature from Fahrenheit to Celsius.

To convert Fahrenheit to Celsius, we use the formula:
°C = (°F - 32) / 1.8

Given:
Initial temperature (Ti) = 71 °F
Final temperature (Tf) = 39 °F

Converting the temperatures to Celsius:

Ti = (71 - 32) / 1.8 = 21.67 °C
Tf = (39 - 32) / 1.8 = 3.89 °C

Now, we can calculate the heat energy (q) transferred from the drink using the formula:

q = mass × specific heat capacity × change in temperature

Given:
Mass of the drink = 13.0 oz
Heat capacity of the drink = 4.18 J/g°C
Change in temperature (ΔT) = Tf - Ti

Converting the mass of the drink from ounces to grams:

1 oz = 28.35 grams

Mass of the drink = 13.0 oz × 28.35 g/oz ≈ 368.55 g

Substituting the values into the formula:

q = 368.55 g × 4.18 J/g°C × (3.89 °C - 21.67 °C)

Simplifying the equation:

q = 368.55 g × 4.18 J/g°C × (-17.78 °C)

Now, we need to convert the heat energy from joules to kilojoules:

1 kJ = 1000 J

q = (368.55 g × 4.18 J/g°C × (-17.78 °C)) / 1000

Simplifying further:

q = -26.24692 kJ

Since heat is transferred from the drink to the ice, the value of q is negative.

Next, we need to calculate the amount of ice required to absorb this amount of heat energy.

Given:
Heat absorbed by 1 gram of ice = 0.33 kJ/g

To find the mass of ice required, we can rearrange the equation as follows:

Mass of ice = q / (heat absorbed by 1 gram of ice)

Mass of ice = -26.24692 kJ / (-0.33 kJ/g)

Simplifying the equation:

Mass of ice ≈ 79.52 g

Therefore, approximately 79.52 grams of ice are required to cool the 13.0 oz drink from 71 °F to 39 °F, assuming 100% heat transfer efficiency.