Question

If 15.0 g of Ice at -25.0°C is mixed with a mass of antifreeze initially at  -3.00°C. If the final temperature of the mixture is -10.0°C, how much  antifreeze is present (Cantifreeze=2.24 x 102 J/KgoC)?

Answer 1

$$m_{ice}c_{ice}(t_f-t_{ice})=m_{antfr}c_{antfr}(t_{antfr}-t_f),$$ $$m_{antfr}=\dfrac{m_{ice}c_{ice}(t_f-t_{ice})}{c_{antfr}(t_{antfr}-t_f)},$$ m_{antfr}=\dfrac{0.015\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot(-10^{\circ}C-(-25^{\circ}C))}{224\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot(-3^{\circ}C-(-10^{\circ}C))}, $$m_{antfr}=0.3\ kg=300\ g.$$