so I have posted this before but this is with some wrok..
Heres the question...
When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 12.0 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 think you do...
first convert 75 and 35 to celsius
then u do 2820*4.18*22.22 watever u get then divide it by 330
can someone please just do the problem and get an answer then round to the correct number of sig figs and tell me? because i have tried 6 times and have gotten it wrong and i only have one more submission!
thanks:)
This is a different question than my sisters if you saw her post
Where dd your 2820 come from?
40F of cooling is 22.22 deg C. You got that part right.
12 fluid ounces is actually a volume measurement. Each fluid oz is 29.57 cm^3. If the density is the same as water, the mass per fluid oz. is 29.57 g.
Remember that the ice that melts also heats up to the final 35 F (1.67 C). It absorbs another 7 J after melting, heating up from 32F to 35F
The amount of heat you must remove from the liquid drink is
12 * 29.57* 22.22 * 4.18 = 32,960 J
The amount of ice needed to remove that heat is 32,960 J/(330 + 1.67*4.18 J/g)
= 32960/337 = 97.8 g
Keep 2 sig figs and call it 98 g
The idea is sound.
In chemistry, you need to be very careful with accuracy, and show all conversion factors.
(75-35)°F=22.22°C is correct.
1 US fl. oz weighs about 29.57 g (check the conversion factor).
So 12 fl.oz (of water) weighs about 354.88 g.
Do not forget that when ice melts, it stays at 0° and eventually heats up to 35°F. The resulting heat exchange has to be accounted for. This will add an extra term to your equation.
Give it another try and post what you've got.
I think i need to change to oz. to ml and then use d=m/v density of water being 1 and find the mass??
Correct.
specific gravity of water = 1
density of water = 1 g/cc (approx.)
It is important (especially in chemistry) to specify units, and to show all the steps how you got your answers.
Someone may need to reproduce your results years later, and giving all details help.
ok so... heres what i did. I am doing a practice problem so just one number is different. instead of 12 oz its 16.0.
.33kj = 300j
16 oz to ml= 473.176475mL
I think that is also the mass except grams because that times 1 is itself.
then i have 22.22222222 for the difference in temp. that's in degrees celcius.
where do i go from here?i multiplied the grams by 4.18 j/g degrees celcius, and that by the temp change.
what do i do?
I got 43952.83701 J but that's not the answer because it is saying how much ice
I Got It!!!!! Thanks:):) woooooo im soo happy
".33kj = 300j"
Did you use 300j ou 330j?
Sure, I can help you with that. Let's break down the problem step by step.
1. Convert the temperatures from Fahrenheit to Celsius:
To convert Fahrenheit to Celsius, we use the formula: Celsius = (Fahrenheit - 32) / 1.8.
So, for the initial temperature of 75°F, we have:
Celsius = (75 - 32) / 1.8 = 23.9°C.
And for the final temperature of 35°F, we have:
Celsius = (35 - 32) / 1.8 = 1.7°C.
2. Calculate the heat absorbed by the drink:
The heat capacity of the drink is given as 4.18 J/g·°C. We are given a 12.0-fluid ounce drink, but we need the mass in grams for the calculation.
Since the density of water is about 1 g/mL, we can assume that the drink has a similar density. Therefore, 12.0 fluid ounces is equivalent to 355.65 mL (you can convert fluid ounces to milliliters using an online converter or multiplication). And since 1 mL of water is approximately 1 gram, the mass of the drink is 355.65 grams.
The heat absorbed by the drink can be calculated using the formula: q = m * c * ΔT, where q is the heat, m is the mass, c is the heat capacity, and ΔT is the change in temperature.
Using the values we have:
q = (355.65 g) * (4.18 J/g·°C) * (1.7°C - 23.9°C) = -3282.3 J.
Note that the final temperature is lower than the initial temperature, so the change in temperature is negative. Therefore, the heat absorbed by the drink is -3282.3 J.
3. Convert the heat absorbed to kilojoules:
We know that when ice melts, it absorbs 0.33 kJ per gram. So, we need to convert the heat absorbed from joules to kilojoules.
1 kJ (kilojoule) = 1000 J (joules). Therefore, we can calculate the heat absorbed in kilojoules as follows:
-3282.3 J * (1 kJ / 1000 J) = -3.2823 kJ.
Note that the negative sign indicates that heat is absorbed by the drink.
4. Calculate the amount of ice required:
We are required to find the amount of ice required to cool the drink. Since we know that ice absorbs 0.33 kJ per gram, we can use the equation:
Amount of ice = Heat absorbed / Heat absorbed per gram of ice
Amount of ice = -3.2823 kJ / 0.33 kJ/g ≈ -9.94 grams.
Since mass cannot be negative, we can round the result to the appropriate number of significant figures: 10 grams.
Therefore, approximately 10 grams of ice are required to cool the 12.0 fluid ounce drink from 75°F to 35°F.
Remember to double-check the calculations and make sure to use the correct conversion factors and equations while solving the problem.