So, using the formula:
[tex]Q=mc\Delta T[/tex]We could replace the values of the question:
[tex]7420J=1.3Kg\cdot c\cdot2.36C[/tex]Given that "c" is the specific heat of the substance, we could find it solving the previous equation:
[tex]c=\frac{7420J}{1.3Kg\cdot2.36C}=\frac{2418.51J}{Kg\cdot C}[/tex]If we look, this specific heat corresponds likely to Glycerine. That's because glycerine has a specific heat equals 2410J/KgC , and this value is extremely similar to the value that we found.
