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

`T_(1)= 30^(@) , m_(2) = 250 g, c_(2) = 0.4 " kcal/lkg."^(@)C= 0.4."cal/g"^(@)C, T_(2)=30^(@)C`
` m_(3) = 10g,T_(3) = 0^(@)C, ` L = 80 cal/g c, (water) = ` 1 "cal/g."^(@)C, T= ? `
` Q_(1)` ( heat lost by calorimeter) =` m_(1)c_(2) (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_(1)T + 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_(1)c_(2)T_(2) = m_(3)L + (m_(1)c_(1) + m_(2)c_(2) + m_(3)c)T`
` 100 g xx 0.1 "cal/g"^(@)C xx 30^(@)C + 250 g xx 0.4 "cal/g".^(@)C xx 30^(@)C`
` 10 g xx 80 "cal/g" + (100 g xx 0.1 "cal/g."^(@)C + 250 g xx 0.4 "cal/g".^(@)C + 10 g 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.