• # question_answer Steam at $100{}^\circ C$ is passed into 20 g of water at$10{}^\circ C$. When water acquires a temperature of $80{}^\circ C,$ the mass of water present will be [Take specific heat of water $=1\,\,cal\,\,{{g}^{-1}}{{\,}^{o}}{{C}^{-1}}$ and latent heat of steam $=540\,\,cal\,\,{{g}^{-1}}$]                                                 [NEET 2014] A)  24 g                 B)       31.5 g C)  42.5 g              D)       22.5 g

 Heat lost by steam = Heat gained by water Let M amount of heat converts into water. $m'\times L=ms\,\Delta t$ $m'\times 540=20\times 1\times (80-10)$ $m'=\frac{20\times 70}{540}=2.5g$ Now, net water $=20+2.5=22.5\,\,g$