Stream at `100^(@)C` is passed into 20 g of water at `10^(@)C`. When water acquires a temperature of `80^(@)C`, the mass of water present will be [Take specific heat of water `=1 cal g^(-1) .^(@)C^(-1)` and latent heat of steam `=540 cal g^(-1)`]

Options

A. 24 g
B. 31.5 g
C. 42.5 g
D. 22.5 g

Answer

D

Explanation

Heat gained by water ` = Delta Q_("gain") = m_w C_w Delta T = 20 xx 1 xx (80 - 10) = 1400 cal`
Heat lost by steam ` =Delta Q_("loss") = m_s L_s + m_s C_w (100 - 80)`
`Delta Q_("gian") = Delta Q_("loss")`
` 1400 = m_s (540) + m_s (20)`
`m_s = 1400/560 = 2.5 g`
`M_("net") = m_w + m_s = 20 + 2.5 = 22.5 gm`