Question

4. A piece of a metal weighing 140 g is initially at 23.0oC and then absorbs 585 cal of heat. The metal has a specific heat of 0.0200 cal/g-1oC-1. What is the final temperature of the metal piece? Assume that all the heat goes to heating the metal.

Answer 1

The piece of metal will heat up according to the equation

$$Q=cm(t_2-t_1),\\ \space\\ t_2=\frac{Q}{cm}+t_1=\frac{585}{140\cdot0.02}+23=232^\circ\text{C}.$$