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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