Question

a 1.50 x10 squared -g piece of glass at a temprature of 70.0 c is placed in a container with 1.00 x 10 squared g of water initially at a temprature of 16.0 c what is the equilibrium temprature of the water

Answer 1

Heat received is the heat lost:

$$c_gm_g(t_f-t_{ig})=c_wm_w(t_f-t_{iw}),\\\space\\ t_f=\frac{c_wm_wt_{iw}+c_gm_gt_{ig}}{c_gm_g+c_wm_w},\\\space\\ t_f=\frac{4200·0.1·16+670·0.15·70}{4200·0.1+670·0.15},\\\space\\ t_f=26°\text C.$$