Question #7ca3b

1 Answer
A08
Mar 1, 2017

Let s be specific heat capacity of the unknown substance.
Heat Lost by this substance when put in calorimeter is found out the following expression
Q=msDelta t .....(1)
Q_"lost"=5.00xxsxx(80.0-45.0)=175s" J" .......(2)

Similarly using (1) to find out Heat gained by water and calorimeter combination
Q_"gained"=Q_"water"+Q_"calorimeter"
Q_"gained"=30.0xx4.1813xx(45-20)+10.0xx(45-20)
Q_"gained"=3136+250=3386" J" .....(3)

Using Law of conservation of energy and equating (2) with (3)
Q_"gained"=Q_"lost"
3386=175s
=>s=19.35" J(g"^@"C")^-1

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