Question

A 52.19 g sample of a substance is initially at 21.3 °C. After absorbing 2.70 × 102 cal of heat, the temperature of the substance is 149.0 °C. What is the specific heat (SH) of the substance?

Answer 1

Hello again @jrigo

We are going to be using the very same equation as before:
\[\Large Q=m \times c \times \Delta T\]
This time I solve the correct variable for you, that is the specific heat \(c\):
\[\Large c=\frac{ Q }{ m \times \Delta T }=\frac{ Q }{ m \times (T_{2}-T_{1}) }\]
Plug in the data and evaluate. If you need additional help don't be afraid to let me know! :)

Answer 2

Same as before, if you like me to check your answer I'll be more than happy to.
Else good luck!