A student is preparing to perform a series of calorimetry experiments. She first

Question

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the Heat Capacity (C_cal) for her coffee cup calorimeter. She pours a 100.0 mL sample of water at 100.0 degree C into the calorimeter containing a 100.0 mL sample of water at 25.0 degree C. She carefully records the final temperature of the water as 61.0 degree C. What is the value of Heat capacity (C_cal) for the calorimeter? (Density of the water is 1.00 g ml, specific heat of water is 4.18 J (g middot degree C)

Explanation / Answer

To obtain heat capacity of the calorimeter, we need to obtain the heat lost by hot fluid and the heat gained by cold fluid, the difference between them will be kept by the calorimeter:

Qgained = -Qlost

Qcal + Qcold = -Qhot

For heat gained:

Qcold = m Cp deltaT

Qcold = (100mL * 1g/mL) * (4.18 J / goC) * (61-25)oC

Qcold = 100g * 4.18 J / goC * 36oC = 15,048.00 J

For heat lost:

Qhot = -m Cp deltaT

Qhot = -(100mL * 1g/mL) * (4.18 J / goC) * (61-100)oC

Qhot = 100g * 4.18 J / goC * 39oC = 16302.00 J

Qcal = 16302 J - 15048 J = 1254 J

Now we divide by temperature gained by calorimeter and water to get heat capacity:

Ccal = 1254 J / 36 C = 34.83 J / oC

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