**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...
__________________
~\o/~
*
Last edited by Woody; Apr 16th 2019 at 04:22 PM.
Reason: Missed out "per Kelvin"
* |