plz help
plz help A calorimeter for which mc= 25JK¯ contains 140 g ofa liquid. An immersion heater is used to provide energy at a rate of 40 W for a total time of 4.0 min. The temperature of the liquid increases by 15.8 oc. Calculate the specific heat capacity of the liquid. State an assumption made in reaching this result.

Dear Student,

Heat Supplied=Heat gained by calorimeter + heat gained by liquid.Thus,Heat supplied=McalorccT+MlclTor, cl=Heat supplied-McalorccTMlTT=15.8°C 15.8°KHeat supplied=40 J/s×4×60 s=9600 Jor, cl=9600-395140×10-3×15.8cl=4161.4 J/Kg/K 4.2 kJ Kg-1K-1

We assume no heat is lost to the surrounding

Regards

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