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:)
8 answers
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
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.
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.
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
.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?
Did you use 300j ou 330j?
0 answers