ok i know i posted this before but it got totally confusing and its late and its due tom.
When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 14.0 fluid ounce drink from 75°F to 35°F, if the heat capacity of the drink is 4.18 J/g·°C? (Assume that the heat transfer is 100% efficient.)
i got 212.56 and rounded it to 213 but it said it was wrong. i need help!
Show your work and I'll find the error.
ok.. i got
(miX330j/g)+(396.893X4.148X+40)+0
70145.28+330j/g(mi)
70145.28=330j/g(mi)
70145.28/330
mi=212.56
Three errors and two of them compensated for each other.
First,
(mass ice x 330) + (mass water x specific heat water x (Tfinal-Tinitial) = 0
1. You reversed the Tfinal and Tinitial. That should be 35-75 = -40 and not +40.
2. When you moved the 70145.28 to the other side you did not change the sign; however, mistake #2 canceled the #1 error.
3. If I multiply 396.893 x 4.184 x -40 I don't get 70,000 something but more like 66,000 something which goes on to obtain slightly more than 200 g ice.
i got 201.28 and rounded to 201 but it said it was wrong
Two possibilities. The other problem I worked had 12 ounces of water and not 14. Could this be the problem?
Second, you are allowed only two places (because of the 40 (assuming the problem gave T as 75 and 35). If the 14 fluid oz is correct, then the correct answer will be 2.0 x 10^2 g ice required.
14 is correct. i tried the 2.0 X 102 and it was wrong. I have no idea what im doing wrong. This is a though question
I have checked my calculations and I stick by the 2.0 x 10^2 grams. I assume you are typing the answer into a database. If you leave out the caret (the ^) the think the database will not put it in for you. Type in 2.0 x 10^2.
in the database im using we have to put 2.0e2 i have tried that but to no evail.
Actually now that look at the prob. again. I think the 40 degrees F. has to be converted to degrees C. so i tried that which makes it 22.22 and put that instead of 40 and got 31.6 grams but it still said it was wrong
Sarah--I'm sorry I didn't catch that. Actually, I didn't catch two or three things. You are right, the F must be converted to C. Also I didn't catch that the ice, which melts at zero C must ALSO be heated to 35 F (converted to C) AND I didn't convert fluid ounces to grams correctly.
To solve this problem, you need to calculate the amount of heat energy required to cool the drink, and then use the heat capacity and the heat of fusion of ice to determine the amount of ice needed.
Step 1: Calculate the amount of heat energy required to cool the drink.
To find the amount of heat energy, you can use the equation:
Q = m * c * ΔT
Where:
Q is the heat energy (in Joules)
m is the mass of the drink (in grams)
c is the specific heat capacity of the drink (in J/g·°C)
ΔT is the change in temperature (in °C)
First, convert the volume of the drink from fluid ounces to grams. Since the density of water is close to 1 g/mL, you can assume that 1 fluid ounce is equivalent to 29.57 grams.
Therefore, 14.0 fluid ounces is equal to 14.0 * 29.57 grams = 413.98 grams.
Next, calculate the change in temperature:
ΔT = final temperature - initial temperature
= 35°C - 75°C
= -40°C
Now you can calculate the amount of heat energy:
Q = 413.98 g * 4.18 J/g·°C * (-40°C)
Q ≈ -69,125.07 J (Note: The negative sign indicates that heat is being removed from the drink)
Step 2: Calculate the amount of ice needed.
Since ice absorbs 0.33 kJ per gram when it melts, you need to convert the heat energy from Joules to kilojoules:
-69,125.07 J ÷ 1000 = - 69.12507 kJ
To convert the energy to grams of ice, use the equation:
Q = m * ΔH
Where:
Q is the heat energy (in kilojoules)
m is the mass of ice (in grams)
ΔH is the heat of fusion of ice (0.33 kJ/g)
Rearranging the equation to solve for m:
m = Q ÷ ΔH
Substituting the known values:
m = -69.12507 kJ ÷ 0.33 kJ/g
m ≈ -209.47 g (Note: The negative sign is due to the energy being removed from the drink)
However, mass cannot be negative, so you should take the absolute value:
m ≈ 209.47 g
Therefore, approximately 209.47 grams of ice are required to cool the drink.
Based on your calculations, you rounded the answer to 213 grams, which is slightly higher than the correct value.