When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 10.0 oz drink from 77 ∘F to 40 ∘F, if the heat capacity of the drink is 4.18 J/g∘C? (Assume that the heat transfer is 100 % efficient.)
Express your answer using two significant figures.
You must know the temperature of the ice before this problem can be worked. It may be 0oC but it could be anything below that.
To find out how much ice is required to cool the drink, we need to determine the amount of heat that needs to be absorbed by the ice.
First, let's convert the temperature difference from Fahrenheit to Celsius:
77 °F - 40 °F = 37 °F
To convert from Fahrenheit to Celsius, we need to use the formula:
°C = (°F - 32) * (5/9)
So, converting the temperature difference:
37 °F * (5/9) = 20.56 °C
Next, we need to calculate the heat energy required to cool the drink. We can use the formula:
Q = mcΔT
Where:
Q is the heat energy required
m is the mass of the drink
c is the specific heat capacity of the drink
ΔT is the temperature change
Given:
Mass of the drink (m) = 10.0 oz
Specific heat capacity of the drink (c) = 4.18 J/g°C
Temperature change (ΔT) = 20.56 °C
First, convert the mass of the drink from ounces to grams:
1 oz = 28.35 g
Mass of the drink (m) = 10.0 oz * 28.35 g/oz = 283.5 g
Now, we can calculate the heat energy required (Q):
Q = (283.5 g) * (4.18 J/g°C) * (20.56 °C)
Q = 23547 J
Now, we need to convert the heat energy from joules to kilojoules:
1 kJ = 1000 J
Heat energy in kilojoules = 23547 J / 1000 = 23.55 kJ
Finally, to determine the amount of ice required to absorb this amount of heat energy, we need to convert the heat absorbed by the ice per gram to kilojoules:
0.33 kJ/g
Divide the total heat energy required by the heat absorbed by the ice per gram:
Ice required in grams = 23.55 kJ / 0.33 kJ/g = 71.36 g
Expressing the answer with two significant figures, the amount of ice required to cool the drink is approximately 71 g.