Answer
Therefore the specific heat capacity of the liquid is 2.09 J/g$^{\circ}$C.
Work Step by Step
We will use the equation $q=mc\Delta t$
q = heat transferred
m = mass
c = specific heat
$\Delta t $ = change in temperature
plug in the values.
47.5 = (13.2) (c) (1.72)
***We multiply 13.2 times 1.72 and then divide 47.5 by that answer
*** solve using a calculator
c= 2.09
Therefore the specific heat capacity of the liquid is 2.09 J/g$^{\circ}$C .