A copper calorimeter of mass `1.5 kg` has `200 g` of water at `25^(@)C`. How much water (approx.) at `50^(@)C` is required to be poured in the calorimeter, so that the equilibrium temperature attained is `40^(@)C`? (Sp. Heat capacity of copper `= 390 J//kg-^(@)C`)

Options

A. `510 g`
B. `450 g`
C. `370 g`
D. `610 g`

Answer

A

Explanation

Let `'m`' g be the required mass of water
Then heat lost by hot water
`Q_("lost") = (mg)(1(cal)/(g-^(@)C))(50-40)^(@)C` ...(i)
Heat gained by calometer
`Q_(1)[(1500g)((390)/(4.2xx10^(3))(cal)/(g-^(@)C))](40-25)^(@)C`
Heat gained by the water present in it
`Q_(2) = [(200g)(1(cal)/(g-^(@)C))](40-25)^(@)C`
Total heat gained by the calorimeter and the water present in it,
`Q_("gain") =Q_(1)+Q_(2)`..(ii)
From principle of calorimetry,
heat lost by hot body = heat gained by cold body
so, equating Eqs (i) and (ii) and solving for `m`, we get
`m=508.93 g`.