解:(1)水吸收的热量$Q_{水吸}=c_{水}m_{水}(t_{水}-t_{0水}) = 4.2\times10^{3}\text{ J/(kg}\cdot^{\circ}\text{C})\times2\text{ kg}\times(60^{\circ}\text{C}-30^{\circ}\text{C}) = 2.52\times10^{5}\text{ J}$
(2)煤油吸收的热量$Q_{煤油吸}=Q_{水吸}=2.52\times10^{5}\text{ J},$由$Q_{吸}=cm\Delta t$得,煤油温度的升高值$\Delta t_{煤油}=\frac{Q_{煤油吸}}{c_{煤油}m_{煤油}}=\frac{2.52\times10^{5}\text{ J}}{2.1\times10^{3}\text{ J/(kg}\cdot^{\circ}\text{C})\times2\text{ kg}} = 60^{\circ}\text{C},$煤油的末温$t_{煤油}=t_{0煤油}+\Delta t_{煤油}=30^{\circ}\text{C}+60^{\circ}\text{C}=90^{\circ}\text{C}$