2) To calibrate your calorimeter cup, you first put 53 mL of cold water in the c

Question

2) To calibrate your calorimeter cup, you first put 53 mL of cold water in the cup, and measure its temperature to be 21.1 °C. You then pour 41 mL of hot water, temperature 53.7 °C, into the cup and measure the temperature every 30 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 34.4 °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 4.184 J/g-K or J/g-°C.) What is the heat change of the cold water, qcw? What is the heat change of the calorimeter cup, qoup?

Explanation / Answer

Since water density = 1gm/cm^3 or 1 gm/mL

Hence we can directly use volume as grams

Density = Mass/Volume, Mass = Volume * Density = 41 mL * 1 gm/mL = 41 grams

Heat change of the hot water = mass of hot water * specific heat capacity * change in temperature

=> 41g * 4.184 J/gC * (34.4-53.7)

=> -3310.7992J

Heat change of the cold water = mass of cold water * specific heat capacity * change in temperature

=> 53g * 4.184 J/gC * (34.4-21.1)

=> 2949.3016J

Heat change of the calorimeter + Heat change of cold water + Heat change of hot water = 0

Heat change of caloriemter = 3310.7992 - 2949.3016 = 361.4976J

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