## Conceptual Physics (12th Edition)

a. Find the amount of heat needed for steel. $$Q = cm \Delta T$$ $$=\frac{450 J}{kg \cdot ^{\circ}C}(10 kg)(100^{\circ}C - 0^{\circ}C ) = 450,000 J$$ b. Now perform the calculation for water. $$Q = cm \Delta T$$ $$=\frac{4190 J}{kg \cdot ^{\circ}C}(10 kg)(100^{\circ}C - 0^{\circ}C ) = 4,190,000 J$$