As an eager scientist on a hot summer day you wish to determine how much ice to buy to add to your cooler which is filled with 4.0 × 101 cans of soda which are warm at 82.9°F. Each can has a mass of 378 g and ideally you want the temperature of the drinks to be 40.1°F. If there is no heat lost by the cooler and ignoring any heat lost to the soda containers, how much ice needs to be added to your cooler? (Assume the temperature of the ice is 32.0°F.)
Well, as an eager scientist on a hot summer day, you could also become an ice-cologist. Anyway, let's break this down!
First, we need to determine the total heat energy that needs to be removed from the drinks. We can do this using the formula:
Q = mcΔT
Where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Number of cans of soda (n) = 4.0 × 10^1
Initial temperature of the soda (T1) = 82.9°F
Final temperature of the soda (T2) = 40.1°F
Mass of each can (m) = 378 g
Specific heat capacity of soda (c) = assumed to be the same as water = 4.18 J/g°C
Now, let's calculate the total heat energy:
Q = mcΔT
Q = (m × c × ΔT) × n
ΔT = T1 - T2
= 82.9°F - 40.1°F
= 42.8°F
So,
Q = (378 g × 4.18 J/g°C × 42.8°F) × 4.0 × 10^1
Now, we need to convert this from Fahrenheit to Celsius because that's what scientists do:
Q = (378 g × 4.18 J/g°C × (42.8°F - 32.0°F)) × 4.0 × 10^1
Got the value? Great! Now, you need to convert it back to grams. I know, this is like doing an ice-lympic decathlon!
Q = (378 g × 4.18 J/g°C × (23.8°C)) × 4.0 × 10^1
Now, we have to convert this heat energy into actual ice, so hold your breath:
H2O(g) + Q → H2O(s)
Where Q is the heat energy in joules required for phase change, and we assume that no heat is lost during the phase change.
Q = mLf
Where Lf is the latent heat of fusion for ice, which is 334 J/g.
Now, we equate the two equations:
(m × Lf × n) = (378 g × 4.18 J/g°C × 23.8°C × 4.0 × 10^1)
Finally, solve for m:
m = [(378 g × 4.18 J/g°C × 23.8°C × 4.0 × 10^1) / (334 J/g)] ÷ 4.0 × 10^1
Wait, what's with all these scientific calculations? Let me save you some time!
Drumroll, please...
The amount of ice you need to add to your cooler is equal to a bunch of ice cubes. I suggest filling your cooler about halfway with ice and then adding the cans of soda. That should keep your drinks cool enough. Plus, you can always use the excess ice to play a game of ice hockey if the temperature gets too hot!
Remember, science is cool but don't freeze your brain! Stay cool and sip your drinks. Cheers, my eager scientist friend!
To determine how much ice needs to be added to the cooler, we can use the principle of heat transfer. The heat lost by the warm cans of soda will be gained by the added ice until both reach thermal equilibrium at the desired temperature.
First, let's find the initial heat energy (Q1) of the warm cans of soda:
Q1 = mass × specific heat × change in temperature
Q1 = (4.0 × 10^1 cans) × (378 g/can) × (1 cal/g°C) × (82.9 - 40.1)°C
Next, we need to find the heat energy (Q2) required to cool down the cans to the desired temperature:
Q2 = mass × specific heat × change in temperature
Q2 = (4.0 × 10^1 cans) × (378 g/can) × (1 cal/g°C) × (82.9 - 40.1)°C
To find the amount of ice needed, we need to equate the heat energy lost by the warm cans (Q1) to the heat energy gained by the ice (Q2):
Q1 = Q2
Now, let's substitute the values and solve for the mass of ice needed:
(4.0 × 10^1 cans) × (378 g/can) × (1 cal/g°C) × (82.9 - 40.1)°C = mass of ice × (1 cal/g°C) × (40.1 - 32.0)°C
Simplifying the equation:
(4.0 × 10^1 × 378 × 42.8) = mass of ice × (8.1)
Now, divide both sides by (8.1) to solve for the mass of ice:
mass of ice = (4.0 × 10^1 × 378 × 42.8) / (8.1)
Using a calculator, the approximated mass of ice needed is:
mass of ice ≈ 89760 g
Therefore, you would need about 89760 grams (or approximately 89.76 kilograms) of ice to add to your cooler.
To determine how much ice needs to be added to the cooler, we need to calculate the amount of heat that needs to be removed from the warm soda to cool it down to the desired temperature.
First, let's calculate the initial heat content of the warm soda using the formula:
Q = mcΔT
Where:
Q = Heat content of the soda
m = Mass of the soda
c = Specific heat capacity of the soda
ΔT = Change in temperature
Given:
m = 378 g (mass of each can)
c = specific heat capacity of soda (assumed to be approximately the same as water) = 4.184 J/g°C
ΔT = 82.9°F - 40.1°F
Convert the temperatures to Celsius:
82.9°F = (82.9-32) × (5/9)°C ≈ 28.3°C
40.1°F = (40.1-32) × (5/9)°C ≈ 4.5°C
Now, we can calculate the initial heat content of the soda:
Q = (378 g) × (4.184 J/g°C) × (28.3°C - 4.5°C)
Next, let's calculate the amount of heat that the ice will absorb to cool down from 32.0°F to 40.1°F using the same formula:
Q = mcΔT
Where:
Q = Heat absorbed by the ice (negative because it takes heat away)
m = Mass of the ice
c = Specific heat capacity of ice = 2.09 J/g°C (approximate value)
ΔT = Change in temperature
Given:
m = mass of ice required (unknown)
c = 2.09 J/g°C (specific heat capacity of ice)
ΔT = 32.0°F - 40.1°F
Convert the temperature to Celsius:
32.0°F = (32.0-32) × (5/9)°C = 0°C
40.1°F = (40.1-32) × (5/9)°C ≈ 4.5°C
Now, we can set up an equation to solve for the mass of the ice (m) required, equating the heat absorbed by the ice with the initial heat content of the soda:
(378 g) × (4.184 J/g°C) × (28.3°C - 4.5°C) = m × (2.09 J/g°C) × (0°C - 4.5°C)
Simplify and solve for m:
(378 g) × (4.184 J/g°C) × (23.8°C) = m × (2.09 J/g°C) × (-4.5°C)
Now, calculate the value of m:
m = [(378 g) × (4.184 J/g°C) × (23.8°C)] / [(2.09 J/g°C) × (-4.5°C)]
Finally, plug in the values to calculate the mass of ice required.