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Old Apr 16th 2019, 10:47 AM   #11
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A bit tricksy

The first part is fairly easy
cool down 1 kg of tea by 75 K
4180 J / kg / K so 4180 * 75 = 31350 J needs to be transferred to the ice.
(I assume you are familiar with kelvin and centigrade and how they relate to each other)

33400 J/kg to melt ice gives 31350 / 33400 = 0.94 kg of ice (rounded off).

None of the answers match!
What I have forgotten?

We also need to heat the water from the ice up by 5 deg.
This requires 4180 Joules per kg of ice per Kelvin

So let X be the mass of the ice, then:
31350 = (33400*X) + (4180*5*X)
31350 = X*(33400+4180*5)
X=31350/(33400+4180*5)

This works out at about 883 grams.

As is often the case in these examples they have rounded the answer off
(this can be suspected from all the supplied options being quoted to just 2 significant figures).

3 grams more or less is not going to change the temperature by much...
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Last edited by Woody; Apr 16th 2019 at 04:22 PM. Reason: Missed out "per Kelvin"
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Old Apr 16th 2019, 11:03 AM   #12
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Note that as a general rule we won't do your homework for you.
I found that by the time I had explained the problem, it was pretty much finished anyway...
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Last edited by Woody; Apr 16th 2019 at 01:50 PM.
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Old Apr 16th 2019, 11:38 AM   #13
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No, I did not mean to imply that there is a rule against "lol". My point was just that if you are "laughing out loud" about this problem you shouldn't be too surprised to get humorous responses.
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Old Apr 17th 2019, 08:51 AM   #14
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Citris Concentrate

Here is a similar problem ~ thoroughly worked. You can see what steps are involved.

Citris Concentrate

Good Luck, TSH
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Old Apr 20th 2019, 03:47 AM   #15
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Rest assured it wasn't the humour I was surprised at, rather my inability to comprehend what they were trying to say, at all.

As you can see I figured it out, and happily went along with the banter. I will welcome any humour-filled responses - I don't want to be a robot all the time. Cheers

-I
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Old Apr 20th 2019, 04:13 AM   #16
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Thank you for this!

I apologise if the way I structured my post was pegging for straight up answers!! It's definitely not like that D:

The walk through is really appreciated nonetheless! I understand the equations but I'll need to go over the concept over lunch or something. I'll know how to answer this question in the future - many thanks!!
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Old Apr 20th 2019, 04:17 AM   #17
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Oof, that is thorough indeed! Wow - your site is extremely resourceful. I will be utilising this for sure. Thank you for your hard work.
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