To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup,

Question

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 21.7 degree C. You then pour 59 mL of hot water, temperature = 50.1 degree C, into the cup and measure the temperature every thirty seconds over a 10-minute period. You extrapolate this cooling curve back to the time of addition and find that the final temperature after mixing is 36.2 degree C. What is the heat change of the hot water, qHW? (Assume the density of the water is 1.00 g/mL, and remember that the specific heat of water is 4.184J/g-K or J/g degree C.)

Explanation / Answer

49 mL *(1.00 g/mL) = 59 g of water that had a temp change (delta t) of -13.9 degrees

Heat loss =59 g * -13.9 degrees * (4.184 J/g-degrees) = -3431 J

Heat change of the cold water?

Same type of equation, but delta t is + 14.5 degrees and mass is 45 grams

14.5*45*4.184=2730 J

Heat change of cold water is +2730 J.

That means that the cup absorbed the extra heat (3198 J - 2230 J) = 701 J

I am going to guess that the cup had time equalize temperature with the cold water before the hot water was added, so the delta t of the cup was also 13 degrees.

Ccup =701 J / 13 degree C

Ccup = 53.9 J / degree C

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