Type: C
Calorimeter
Heat lost by = Heat gained + Heat gained
hot object = by calorimeter + by liquid
`Q = Q_(1) + Q_(2)`
`Q = mcDeltaT`
Q = mL
Note : `C_(Copper) = 0.1cal//g^(@)C = = 0.1 kcal//kg^(@)C`
Final temperature becomes equal to 'T' for all substances
A copper sphere of of 100g mass in heated to raise its temperature to `100^(@)C` and is released in water of mass 195g and temperature `20^(@)C` in a copper calorimeter . If the mass of calorimeter is 50g what will be the maximum temperature of water ?

Explanation

To find : T = ?
Formuale : `Q = Q_(1) + Q_(2)`
`Q = mcDeltaT`
Solution:
(1) Copper
` Q = mcDeltaT……………(i)`
(2) Water
`Q_(1) = m_(1)c_(1)DeltaT_(1)…………..(i)`
(3) Calorimeter
`Q_(2) = m_(2)c_(2)DeltaT_(2)..........(iii)`
from (i) ,(ii) and (iii)
`mcDeltaT = m_(1)c_(1)DeltaT_(1) + m_(2)c_(2)DeltaT_(2)`
`{100xx0.1xx(100-T)} = {195 xx 1 xx (T - 20)} + {50 xx 0.1 xx (T-20)}`
`10(100-T) = 195(T -20) + 5(T-20)`
`1000-10T = (T - 20) (195 + 5)`
` 1000 - 10T= 200T - 400`
`1000+ 4000 = 200 T + 10T`
`5000 = 210T`
` T = (500cancel0)/(21cancel0) = 23.8^(@)C`
The maximum temperature of water will be `23.8^(@)C`