A calorimeter has mass 100g 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` , what will be the temperature of the mixture ?

Explanation

To find : T = ?
Formulae : `Q = Q_(1) + Q_(2)` S
`Q = mcDeltaT ,Q = mL`
Solution :
(1) Ice
`Ice(0^(@)C)overset(mL)to "Water"(0^(@)C)overset(mcDeltaT)to"Water"(T^(@)C)`
`therefore Q= mL + mcDeltaT..........(i)`
(2) Water
`Q_(1) = m_(1)c_(1)DeltaT_(1)..........(ii)`
(3)Calorimeter
`Q_(2) = m_(2)c_(2)DeltaT_(2)...........(iii)`
from(i),(ii)and (iii)
`Q= Q_(1) + Q_(2)`
`mL + mcDeltaT = m_(1)c_(1)DeltaT + m_(2)c_(2)DeltaT_(2)`
`(10 xx 80) + {10xx1xx(T -0)} ={250 xx 0.4 xx (30-T)}+{100 xx 0.1xx (30-T)}`
`800 + 10T = 100 (30-T)+ 10(30-T)`
`800 + 10T =110(30-T)`
`10T + 110T = 3300-800`
`120T = 2500`
`T = (250cancel0)/(12cancel0) = 20.8^(@)C`