解：?$(1)Q_{水}=c_{水}m_{水}?t=4.2×10^3\ \mathrm {J/}(\mathrm {kg·℃}) ×0.1\ \mathrm {kg}×\frac 1{12}×(30℃-0℃)=1050\ \mathrm {J}$?
?$Q_{油}=Q_{水}=1050 \text {J}$?
?$(2)c_{油}=\frac {Q_{油}}{m_{油}?t'}=\frac {{ 1050 } \text {J}}{{ 0.1 } \text {kg}×\frac 16(30℃-0℃)}={ 2.1×10^3 } \text {J/}( \text {kg}·℃)$?
