A student adds 110kJ of heat to a calorimeter containing 250 grams of water. The temperature in the calorimeter changes from 23 degrees celcius to 98 degrees celcius. What is the percent efficiency of the calorimeter?
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taking the water as the basis for comparison
Q = m c delta(T)
Q = 250 g (4.186 J/(g*C) ) ( 75 C) = 78500 J or 78.5 kJ
this was the amount absorbed by the water
the rest was absorbed by the calorimeter or lost to the environment
so
X/100 = 78.5/110
X = 71.4%
Q = m c delta(T)
Q = 250 g (4.186 J/(g*C) ) ( 75 C) = 78500 J or 78.5 kJ
this was the amount absorbed by the water
the rest was absorbed by the calorimeter or lost to the environment
so
X/100 = 78.5/110
X = 71.4%