When ice melts, it absorbs 0.33 per gram.
How much ice is required to cool a 11.0 drink from 75 to 37, if the heat capacity of the drink is 4.18 ? (Assume that the heat transfer is 100 efficient.)
You have no units in your numbers, making the problem unsolvable, and meaningless.
To solve this problem, we need to calculate the amount of heat energy that needs to be absorbed by the ice in order to cool the drink.
The formula we can use is:
Q = m * c * ΔT
Where:
Q is the heat energy absorbed or released,
m is the mass of the substance (in this case, the ice),
c is the specific heat capacity of the substance (in this case, the heat capacity of the drink),
ΔT is the change in temperature.
First, let's calculate the change in temperature (ΔT) of the drink:
ΔT = final temperature - initial temperature
ΔT = 37 - 75
ΔT = -38°C
Since the change in temperature is negative, it means heat is being removed from the drink.
Now, let's calculate the amount of heat energy (Q) that needs to be absorbed by the drink:
Q = m * c * ΔT
Rearranging the formula, we can solve for mass:
m = Q / (c * ΔT)
Since we know the heat capacity of the drink (c = 4.18) and the change in temperature (ΔT = -38), we can substitute these values into the formula:
m = Q / (4.18 * -38)
Now, we need to know the value of Q, which is the amount of heat energy that needs to be absorbed by the ice to cool the drink. We can calculate Q using the equation:
Q = mcΔT
Where:
m is the mass of the substance (in this case, the ice),
c is the specific heat capacity of the substance (in this case, 0.33 J/g°C),
ΔT is the change in temperature (in this case, -38°C).
Since we're solving for Q, we can rearrange the formula:
Q = mcΔT
Q = m * 0.33 * -38
Now, we can substitute this value of Q back into the mass formula:
m = Q / (4.18 * -38)
m = (m * 0.33 * -38) / (4.18 * -38)
Now, we can simplify the equation:
m = (-0.33m * 38) /(-38 * 4.18)
Cancelling out the common terms:
m = (-0.33) / (4.18)
And further simplifying:
m ≈ -0.079
Since the mass cannot be negative, we can conclude that approximately 0.079 grams of ice is required to cool the drink from 75 to 37 degrees.