Question

A 223 g block of iron that is 70◦C is placed in some volume of water at 20◦C. The final temperature of the iron and the water is 45◦C. What is the mass of the water in grams. Use the specific heat capacity of iron to be 0.444 J/g◦C and the specific heat capacity of water is 4.186 J/g◦C.

Answer 1

$$m_{iron}c_{iron}(t_{iron}-t_f)=m_{w}c_{w}(t_{f}-t_{w}),$$ $$m_{w}=\dfrac{m_{iron}c_{iron}(t_{iron}-t_f)}{c_{w}(t_{f}-t_{w})},$$ m_{w}=\dfrac{223\ g\cdot0.444\ \dfrac{J}{g\cdot \!^{\circ}C}\cdot(70^{\circ}C-45^{\circ}C)}{4.186\ \dfrac{J}{g\cdot \!^{\circ}C}\cdot(45^{\circ}C-20^{\circ}C)}, $$m_{w}=23.65\ g.$$