A calorimeter has mass 100 g and specific heat 0.1 kcal/`kg.^(@)C`. It contains 250 g of liquid at ` 30^(@) C ` having specific heat of 0.4 kcal/ ` kg.^(@)`C . If we drop a piece of ice of mass 10 g at ` 0^(@) c` into it , what will be the temperature of the mixture ?

Explanation

Data `:m_(1) =100g, c_(1)= 0.1 kcal //kg.^(@)C , 0.1 cal //g.^(@)C, T_(1) = 30^(@) C ,m_(2)=250 g,c_(2)= 0.4kcal//kg.^(@)C,T_(2) = 30^(@)C , m_(3)=10g, T_(3) = 0^(@)C, L=80cal //g `
c ( water ) =`1 cal // g. ^(@)C, T = ?`
`Q_(1) ` ( heat lost by calorimeter ) = `m_(1)c_(1) (T- T_(1))`, `Q_(2) ` ( heat lost by liquid ) `= m_(2)c_(2) (T-T_(2))`, `Q_(3)` ( heat absorbed by ice ) ` =m_(3) L`,
`Q_(4)` ( heat absorbed by water formed on melting of ice )`=m_(3) c ( T -0^(@) C ) `
According to the principle of heat exchange , `Q_(1) + Q_(2) = Q_(3)+Q_(4)`
`:.m_(1)c_(1)(T_(1)-T)+ m_(2)c_(2)(T_(2)-T) = m_(3)L + m_(3)c ( T - 0^(@)C)`
`:.m_(1)c_(1)T_(1)-m_(1)c_(2)T_(2) +m_(2)c_(2)T_(2) -m_(2)c_(2)T= m_(3)L + m_(3)c(T- 0^(@)C)`
`:. m_(1)c_(1)T_(1) + m_(2)c_(2)T_(2) =m_(3)L = ( m_(1)c_(1)+m_(2)c_(2)+m_(3)c) T `
`:. 100g xx0.1 cal//g.^(@) C xx30^(@) C + 250 g xx 0.4 cal //g.^(@)C xx 30^(@) C `
` 10g xx 80 cal //g +( 100g xx 0.1 cal //g.^(@) C + 250 g xx 0.4 cal//g.^(@) C + 10g xx 1 cal //g.^(@) C ) T `
`:. ( 10 + 100+ 10)T = 300+ 3000 - 800)^(@)C `
`:. 120 T = 2500^(@)C `
`:. T =( 2500)/( 120 ).^(@)C = ( 125)/( 6).^(@)C =20.83^(@) C `
This is the temperature of the mixture.