When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 24.0 fluid ounce drink from 75°F to 35°F, if the heat capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)
Here is my response to the same question to Rebekah.
http://www.jiskha.com/display.cgi?id=1284593123
To determine the amount of ice required to cool the drink, we need to calculate the heat transferred from the drink to the ice.
Here's how you can solve this problem step by step:
Step 1: Convert the volume of the drink from fluid ounces to grams.
To do this, we need to know the density of the drink or assume a value. Let's assume the density of the drink is close to that of water, which is approximately 1 g/mL.
24.0 fluid ounces ≈ 710.87 grams (assuming 1 g/mL)
Step 2: Calculate the change in temperature of the drink.
The initial temperature is 75°F and the final temperature is 35°F.
Change in temperature = Final temperature - Initial temperature
Change in temperature = (35°F - 75°F)
Change in temperature = -40°F
Step 3: Convert the change in temperature from Fahrenheit to Celsius.
To do this, we need to use the conversion formula: °C = (°F - 32) / 1.8
Change in temperature = (-40°F - 32) / 1.8
Change in temperature = -40°C
Step 4: Calculate the heat energy transferred from the drink to the ice.
Heat energy transferred = mass of the drink × specific heat capacity × change in temperature
Heat energy transferred = 710.87 g × 4.18 J/g·°C × (-40°C)
Heat energy transferred = -118,997.59 J
Step 5: Convert the heat energy from joules to kilojoules.
Heat energy transferred = -118,997.59 J ÷ 1000
Heat energy transferred = -118.99759 kJ
Step 6: Calculate the mass of the ice required.
To calculate the mass of the ice, we need to use the heat of fusion of ice, which is the energy required to melt ice.
0.33 kJ of energy is required to melt 1 gram of ice.
Mass of ice required = (-118.99759 kJ) / (0.33 kJ/g)
Mass of ice required ≈ -360.9 grams (Since mass cannot be negative, it means we require 360.9 grams of ice)
So, approximately 360.9 grams of ice is required to cool the 24.0 fluid ounce drink from 75°F to 35°F, assuming 100% efficiency.