While calibrating a coffee cup calorimeter, you add 50 mL of hot water, initiall

Question

While calibrating a coffee cup calorimeter, you add 50 mL of hot water, initially at 60°C, to 50 mL of cold water, initially at 20°C, and find that the final temperature is 24°C. Assume that all of the heat released by the hot water is either absorbed by the cold water or the calorimeter itself. Assume that water has a density of 1.0 g/mL.

What is the heat capacity of the calorimeter?

            A.        -1.67 kJ/°C

            B.         1.67 kJ/°C

            C.         -558 kJ/°C

            D.        558 kJ/°C

Explanation / Answer

we know that

mass = density x volume

given

denisty = 1 g / ml

so

mass of hot water = 50 x 1 = 50 g

mass of cold water = 50 x 1 = 50 g

now

we know that

heat lost by hot body = heat gained by cold body

so

heat lost by hot water = heat gained by cold water + calorimeter

we know that

heat lost by water = mass x specific heat x temp change

and

heat lost by calorimeter = heat capacity x temp change

now

heat lost by hot water = heat gained by cold water + calorimeter

( m x s x dT ) for hot water = ( m x s x dT ) for cold water + ( heat capacity x temp change of calorimter)

50 x 4.184 x ( 60-24) = 50 x 4.184 x ( 24-20) + [ c x (24-20) ]

7531.2 = 836.8 + 4c

c = 1673.6

c = 1.6736 x 1000

so

heat capacity of calorimeter = 1.67 kJ / C

B) 1.67 kJ/C

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