To calibrate your calorimeter, you first put 45 mL of cold water in the cup, and
ID: 967622 • Letter: T
Question
To calibrate your calorimeter, 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, initial 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 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, q_HW? (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- degree C.) What is the heat change of the cold water, q_CW? What is the heat change of the calorimeter, q_cal? What is the heat capacity of the calorimeter, cal?Explanation / Answer
Cold Water => Volume = 45 ml and Temperature = 21.7 °C
Hot Water => Volume = 59 ml and Temperature = 50.1 °C
Final temperature = 36.2 °C
Density of water = 1.00 g/ml
Specific heat of water = 4.184 J / g °C
With the formula: q=mCT
Where:
q= is the heat energy gained or lost by a substance
m= is the mass (in grams)
C= specific heat of subtance
T= change in temperature (final temperature – initial temperature)
Grams of hot water: density = mass / volume => mass = density x volume
59 ml x (1.00 g/ml) = 59 grams of hot water
qHW = (59 g) (4.184 J / g °C) (36.2 °C – 50.1 °C)
qHW = - 3428.02 J
Grams of hot water: density = mass / volume => mass = density x volume
45 ml x (1.00 g/ml) = 45 grams of cold water
qHW = (45 g) (4.184 J / g °C) (36.2 °C – 21.7 °C)
qHW = + 2730.06 J
qcal + qHW + qCW = 0
qcal + (- 3428.02 J) + (2730.06 J) = 0 ----> qcal + (-697.96 J) = 0
qcal = + 697.96 J
We calculate this based on the final temperature (36.2 °C) and the initial temperature of the cold water, since we are assuming that the initial temperature of the cup was the same as the cold water (21.7 °C), so, the temperature for the calorimeter is going to be (36.2 °C – 21.7 °C = +14.5 °C).
Cal = qcal / T ---------> Cal = + 697.96 J / (36.2 °C – 21.7 °C)
Cal = + 48.14 J / °C
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