Steam at 100 °C is passed into 20 g of water at 10 °C. When the water acquires a temperature of 80 °C, the mass of water present will be (take the specific heat of water =1 cal g⁻¹ ° C⁻¹ and the latent heat of steam =540 cal g⁻¹)

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(4) 22.5 g. Given: m_w=20 g at 10 °C, final temperature 80 °C, s=1 cal g⁻¹ ° C⁻¹, L=540 cal g⁻¹. Idea: the steam condenses at 100 °C and the water so formed then cools to 80 °C; all of that heat is taken up by the original water. Heat given by m grams of steam: mL+ms(100-80)=540m+20m=560m Heat taken by the water: 20×1×(80-10)=1400 cal 560m=1400 m=2.5 g Mass of water present =20+2.5=22.5 g

Steam at 100 °C is passed into 20 g of water at 10 °C. When the water acquires a temperature of 80 °C, the mass of water present will be (take the specific heat of water =1 cal g⁻¹ ° C⁻¹ and the latent heat of steam =540 cal g⁻¹)

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