## College Physics (7th Edition)

a) (1) more heat. More is the specific heat, more heat is needed. As the specific heat of Cu is more than Pb, copper will require more heat than lead. b) $Q_{Cu}-Q_{Pb}=2.1\times10^{4}\,J$
b) $Q= cm\Delta T$ Then $Q_{Cu}=390\,J/(kg\,^{\circ}C)\times1.0\,kg\times80^{\circ}C=3.1\times10^{4}\,J$ $Q_{Pb}= 130\,J/(kg\,^{\circ}C)\times1.0\,kg\times80^{\circ}C=1.0\times10^{4}\,J$ $Q_{Cu}-Q_{Pb}=3.1\times10^{4}\,J-1.0\times10^{4}\,J=2.1\times10^{4}\,J$