`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

Options

A. 20 g of water
B. 20 g of ice
C. 10g ice and 10 g water
D. 5 g ice and 15 g water

Answer

C

Explanation

Heat given of water `(Q_1) = m_w s_w Delta T`
`= 10 xx 1 xx 10 = 100` cal
Heat absorbed by ice `(Q_2) = mL + m_i s_i Delta T`
`= 10 xx 80 + 10 xx 0.5 xx 20`
`= 800 + 100 = 900 cal`
Since, `Q_1 lt Q_2` (Ice will not completely melt)
Heat supplied by water is only just sufficient to raise the temperature of ice from `-20^@ C` to `0^@ C`, so mixture will be of g 10 ice and g 10 water