Question

I don't know why the last two are wrong.

To calibrate your calorimeter cup, you first put 47 mL of cold water in the cup, and measure its temperature to be 25.9 °C. You then pour 49 mL of hot water, temperature = 49.9 °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 37.2 °C. What is the heat change of the hot water, Chw? (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.) -2604.J You are correct. Previous Tries What is the heat change of the cold water, acw? 2222. J You are correct. Previous Tries What is the heat change of the calorimeter cup, Acup? 382J What is the heat capacity of the calorimeter cup, Ccup? (Enter units either as J/K or J/degC). .352J/degc Submit Answer Incorrect. Tries 2/99 Previous Tries

Answer 1

The heat of the calorimeter is calculated:

q cal = 2222 - 2604 = - 382 J

Ccal is calculated:

Ccal = 382 / (37.2 - 25.9) = 33.81 J / ° C

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