(1)由P=UI得,I= P U = 1W 22V = 1 22 A,串联电阻的电压为:U'=U总-U=220V-22V=198V,串联电阻的阻值为:R= U′ I = 198V 1 22 A =4356Ω.(2)灯正常工作10min,电阻工作10min,电阻产生的热量:Q=I2Rt=( 1 22 A)2×4356Ω×10×60s=5400J.故答案为:4356Ω;5400J.