When ice melts, it absorbs 0.33 kJ per gram.
How much ice is required to cool a 11.0 oz drink from 75 degree Fahrenheit to 37 degree Fahrenheit, if the heat capacity of the drink is 4.18 J/g degree celsius ? (Assume that the heat transfer is 100 % efficient.)
Heat givenoff=masswater*specificheat*deltaTEmp
heat to melt ice= massice*Hf
set the heat to melt ice=heatgiven off
and solve for themass of ice.
watch units.
84
To solve this problem, we need to calculate the amount of heat energy required to cool the drink and then convert it to the amount of ice needed.
First, we need to calculate the change in temperature of the drink.
From 75 degrees Fahrenheit to 37 degrees Fahrenheit, the change in temperature is:
ΔT = Final temperature - Initial temperature
ΔT = 37°F - 75°F
ΔT = -38°F
Next, we need to convert this temperature change to degrees Celsius.
To convert Fahrenheit to Celsius, we use the formula:
°C = (°F - 32) * (5/9)
So, the change in temperature in degrees Celsius is:
ΔT = (-38°F - 32) * (5/9)
ΔT = -70°C
Now, we can calculate the amount of heat energy required to cool the drink.
The formula for heat energy is:
Q = m * C * ΔT
Where:
Q is the heat energy,
m is the mass of the substance,
C is the specific heat capacity, and
ΔT is the change in temperature.
In this problem, we are given the heat capacity of the drink (C) as 4.18 J/g°C.
We need to convert the mass of the drink from ounces to grams.
1 ounce = 28.35 grams
So, the mass of the drink in grams is:
m = 11.0 oz * 28.35 g/oz
m = 311.85 g
Now, we can calculate the heat energy required to cool the drink:
Q = (311.85 g) * (4.18 J/g°C) * (-70°C)
Q = -90786.6 J
Since the heat transfer is assumed to be 100% efficient, the heat energy required to cool the drink is equal to the heat energy absorbed by the ice.
Finally, we can convert the heat energy absorbed by the ice to the mass of ice using the given information that ice absorbs 0.33 kJ per gram.
To convert J to kJ, we divide by 1000:
-90786.6 J / 1000 = -90.7866 kJ
To calculate the mass of ice:
m = Q / (0.33 kJ/g)
m = -90.7866 kJ / (0.33 kJ/g)
m ≈ -275.41 g
Since we cannot have a negative mass, the negative sign indicates the direction of heat flow. Therefore, to cool the drink from 75°F to 37°F, we would need approximately 275.41 grams of ice.