A steam at `100^@C` is passed into 1 kg of water contained in a calorimeter of water equivalent 0.2 kg at `9^@C` till the temperature of the calorimeter and water in it is increased to `90^@C`. The mass of steam condensed in kg is nearly (specific heat of water =`1 cal//g^@C`,latent heat of vaporisation= 540 cal`//`g)

Options

A. 0.81
B. 0.18
C. 0.27
D. 0.54

Answer

B

Explanation

At steady state condition of calorimeter,
Heat lost by steam + Heat lost by water = Heat gained by calorimeter
(i) Heat lost by steam ` = m_(s) xx L = m_(s) xx 540 cal//g`
(ii) Heat lost by water `= m_(w) xx S_(w) xx (T_(i) - T_(f))_(w)`
`= 1 xx1 xx (100 - 90) cal//g`
(iii) Heat gained by calorimeter `= (m_(w) + m_(s)) xx S_(c) xx (T_(f) - T_(i))_(c)`
`= (1000 + 200) xx 1 xx (90-9) cal//g`
`therefore` `(m_(s) xx 540) + 10 = 1200 xx 81 = 179.981 g (or) 0.179 kg`