The water equivalent of a calorimeter is 10 g and it contains 50 g of water at `15^@ `C. Some amount of ice, initially at `-10^@ `C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = `0.5 cal g^(-1) ""^(@) C^(-1),` specific heat of water `= 1.0 cal g^(-1) ""^(@) C^(-1)` and latent heat of melting of ice = 80 cal `g'^(-1)`)?

Options

A. 10 g
B. 18 g
C. 20 g
D. 30 g

Answer

C

Explanation

Let initial mass of the ice be m.

Mass of water equivalent of a calorimeter and water contined by it

`m_1 = 10 g + 50 g =60 g `
the amount of heat lost by the water ` Q_1 = m_1 s Delta T `
where s= specific heat of water ,` Delta T =` temperature
` therefore Q_1 ( 60 xx 1 xx 15 )` cal
amount of heat gained by ice is
`Q_2 =(m )/(2) xx 10 + (m)/(2) xx 80 = (45 m)` cal
in steady state heat gained `= `heat lost
` therefore 45 m = 60 xx 1 xx 15 implies m -20 g`