How much ice is required to cool a 15.0oz drink from 71 F to 34 F, if the heat capacity of the drink is 4.18 J/g*C ? (Assume that the heat transfer is 100% efficient.)
I got a very close answer, I just needed help with the steps, so thank you my friend.
To determine the amount of ice required to cool the drink, we need to calculate the heat transfer involved. We can use the formula:
Q = m × ΔT × C
Where:
Q is the heat transfer in joules (J)
m is the mass in grams (g)
ΔT is the change in temperature in degrees Celsius (°C)
C is the specific heat capacity in J/g°C
First, we need to convert the mass of the drink from ounces to grams. Since 1 ounce is approximately 28.35 grams, the mass of the drink is:
m = 15.0 oz × 28.35 g/oz = 425.25 g
Next, we calculate the change in temperature:
ΔT = 34°C - 71°C = -37°C
Now, we can substitute the values into the formula:
Q = 425.25 g × -37°C × 4.18 J/g°C
Calculating this expression gives us the amount of heat transfer. However, since the problem states that the heat transfer is 100% efficient, we can equate this heat transfer to the heat released by the melting of the ice.
The heat released by the melting of the ice can be calculated using the formula:
Q = m × ΔH_fusion
Where:
Q is the heat transfer in joules (J)
m is the mass in grams (g)
ΔH_fusion is the heat of fusion in J/g
Since we need to find the mass of ice required, we can rearrange the formula as follows:
m = Q / ΔH_fusion
Before we can solve for the mass of ice, we need to know the heat of fusion, which is the amount of energy required to change one gram of ice to liquid water at its melting point. The heat of fusion for water is approximately 333.55 J/g. Substituting this value and the calculated heat transfer (Q) into the rearranged formula, we can solve for the mass of ice (m).
425.243 g oz
26.1111c
0.555556 c 33
1.3794
Convert 15 oz to grams. I get about 425 but you should confirm that.
Convert 761F and 34F to celsius.
(mass ice x heat fusion) + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass warm water x specific heat warm water x (Tfinal-Tinitial)] = 0
Solve for x = mass ice. I ran this through the calculator (just once so you should check everything) and I get something like 108 g ice. Note that this assumes the ice is at a T of zero C.