[Hess\' Law] A student constructs a coffee cup calorimeter and places 50.0 mL of

Question

[Hess' Law]

A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the temperature of the water in the calorimeter is determined to be 20.00 oC. To this is added 50.0 mL of water that was originally a temperature of 54.5 oC. A careful plot of the temperatures recorded after this established the temperature at To was 31.25 oC.

What is the colrimeter constant in J/oC for this calorimeter?

(Density H2O = 1.00 g/mL; specific heat of water = 4.184 J/g*oC)

Explanation / Answer

cold water lost:
dH = m C dT
dH = 50g (4.184J/g-C)(31.25C -20C)
dH = 2353.5 Joules

hot water got:
dH = m C dT
dH = 50 (4.184)(31.25c - 54.5c)
dH = -4863.9 joules

dH cold water = - dH of hot water

the calorimeter got the rest:
4863.9 - 2353.5 = 2510.4Joules

find heat capacity of the calorimeter:
2510.4 Joules / 11.25 C temp rise = 218.2957 Joules / C

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