How to Calculate the Specific Heat Capacity of the Calorimeter?

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SUMMARY

The discussion focuses on calculating the specific heat capacity of a calorimeter containing 70g of oil at 30°C and a 125g brass piece heated to 100°C, resulting in a final mixed temperature of 44°C. The specific heat capacities provided are 1.52 J/g°C for oil and 0.42 J/g°C for brass. The calculations yield 1489.6 J for the oil and 2940 J for the brass. The heat gained by the calorimeter can be determined using the equation Qbrass = Qoil + Qcal, confirming that the temperature change for the calorimeter is indeed 14°C.

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Homework Statement



A calorimeter contains of 70g oils at temperature of 30oC.If a piece of brass with a mass of 125g heat up until 100oC.After that a brass put in that calorimeter and final temperature of the mixed is 44oC.Calculate the specific heat capacity (c) of the calorimeter if its mass is 100g.(Given c of oil =1.52Jg-1oC-1 & c of brass =0.42Jg-1oC-1)

Homework Equations



Q=mcθ

The Attempt at a Solution



Qoil=70(1.52)(14)=1489.6J
Qbrass=125(0.42)(56)=2940J

Then I stuck.How to find temperature & Q of calorimeter?
 
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Qoil=70(1.52)(14)=1489.6J
Qbrass=125(0.42)(56)=2940J

Find heat gained by the calorimeter. And then

Qbrass = Qoil + Qcal.
 
Thanks for the help,but just to make sure,is temperature change for calorimeter=44-30=14?
 
kai92 said:
Thanks for the help,but just to make sure,is temperature change for calorimeter=44-30=14?

Yes. Oil is in the calorimeter. So the change in temperature of oil and the calorimeter is the same.
 

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