Heat is lost by steam in two stages (i) for change of state from steam at ${100}^0\mathrm{C}$ to water at ${100}^0\mathrm{C}$ is $\mathrm{m}\times 540\left(\mathrm{ii}\right) \mathrm{to} \mathrm{change} \mathrm{in} \mathrm{water} \mathrm{at} {100}^0\mathrm{C}$ to water $\mathrm{at} {80}^0\mathrm{C} \mathrm{is} \mathrm{m}\times 1\times (100-80), \mathrm{where} \mathrm{m} \mathrm{is} \mathrm{the} \mathrm{mass} \mathrm{of} \mathrm{steam} \mathrm{condensed}.$

Total heat lost by steam = $\mathrm{m}\times 540+\mathrm{m}\times 20 = 560 \mathrm{m} \mathrm{cal}$

Heat gained by calorimeter and its contents = $(1.1+0.02)\times (80-15) = 1.12 \times 65 \mathrm{cal}$

Using principle of calorimetry,

Heat gained = heat lost

      560 m = 1.12 $\times$65

         m = 0.130 gm