# Chemistry

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On more than one occasion when our kids were small, we would come inside from playing in the snow and I would make hot chocolate by heating the chocolate milk in the microwave. Invariably, I would get the milk too hot for at least one of them. One time, rather than adding more cold milk to the hot chocolate (there wasn’t room in the mug), I dropped a couple of small ice cubes into the mug.if the initial temperature of the hot chocolate was is 175̊F, what would the temperature of the drink be once the ice cubes had completely melted? Express your answer in “degrees Celsius” to one decimal place. Additional information is provided below. For this problem, let’s assume that the initial temperature was 175̊F and the mug contained 325 g of hot chocolate.
Weight of ice cubes = 58 grams total weight
Initial temperature of ice cubes = - 18̊C
Specific heat capacity of the hot chocolate = 3.751 kJ/kgC̊

• Chemistry - ,

First convert 175 F to C. I think that's about79 but you need to do it more accurately.
You have omitted one piece of info; i.e., the specific heat of solid H2O(ice).
heat lost by ice in going from -18C to zero C + heat lost by ice melting at zero C + heat gained by melted ice + heat lost by 175 F chocolate liquid = 0

[mass cubes x specific heat ice x (Tfinal-Tinitial)] + [mass cube x heat fusion] + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass 175 F chocolate x specific heat chocolate x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf which is the only unknown (after you have determined the specific heat of solid ice--that should be in the problem somewhere).
Post your work if you get stuck.

• Chemistry - ,

Hello thanks for the reply that is what I had done but I keep getting a value that does not make sense. I possibly have the negative/positive signs mixed up.

I know that heat is conserved so the heat lost by the milk would equal the heat gained by the ice. In your post you said heat is lost by ice going from -18 to 0 would that not be heat gained?

this is my solution
heat gained by ice from -18 to 0 + latent heat x mass (heat gained by meting the ice at zero) + heat gained from melting the ice from 0 to T2 (unknown final temperature) = - [heat lost from the milk going from 0 to T2)

(0.58kg)(2.050kJ/kgC)[(0-(-18)] + (333kJ/khg)(0.58kg) + (0.58kg)(4.187kJ/kgC)(T2-0) = - (0.325kg)(3.751kJ/kgC)(T2-79.44)

when I rearranged and solved for T2 I got a negative large value which does not make sense.

Any advice? Do I have the concept of heat lost/gained confused?

Thanks

• Chemistry - ,

I completed the math using your solution and it worked I have the concept of heat gained and heat lost confused could you assist with that?

• Chemistry - ,

Sorry for all the posts when I did the math your way I got 97.3 degress Celcius which does not make sense either because that would be higher then the starting temperature :S

• Chemistry - ,

I've gone over this several times and I think you must have just made an error in algebra/math. The numbers you have substituted look ok to me. I worked the problem and obtained 51.57 which would round to 51.6. If we substitute that into each phase of the problem it works ok for me.
As for the concept, of course heat must be gained to raise the -18 C ice to zero C. Here is my final check on the phases but I've changed everything to J/g*C.
-18 to 0 C is 58*2.05*18 = +2140.2 J.
melt ice is 58*333 = +19314 J.
raise melted ice from 0 to 51.57C = 58*4.186*(51.57-0) = +12,520.6 J
lower heat of 79.44C milk to Tfinal is 325*3.751*(51.57-79.44) = -33,975.62 J.
All of that should add to zero.
2140.2 + 19314 + 12,520.6 - 33975.62 = -0.82J which is almost zero J. The small difference is due to rounding all of those large numbers so I think 51.6 C should be correct. One final note here. I used 4.186 J/g*C for the melted ice in raising it from zero to final T; however, I suppose a case could be made for using 3.751 J/g*C. Dumping 58 g H2O (at 4.186) into 325 g chocolate (at 3.751) and the MIXTURE must be very close to 3.751; however, I think the correct theoretical approach is what I've used. Good luck with your studies.

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