When 0.400 kg of water at ${30}^0\mathrm{C} \mathrm{is} \mathrm{mixed} \mathrm{with} 0.150 \mathrm{kg} \mathrm{of} \mathrm{water} \mathrm{at} {25}^0\mathrm{C} \mathrm{contained} \mathrm{in} \mathrm{a} \mathrm{calorimeter},$$\mathrm{the} \mathrm{final} \mathrm{temperature}$$\mathrm{is} \mathrm{found}$ 

$\mathrm{to} \mathrm{be} {27}^0\mathrm{C}$ , find the water equivalent of the calorimeter.

Water of mass ${\mathrm{m}}_1 = 0.150 \mathrm{kg} \mathrm{is} \mathrm{taken} \mathrm{in} \mathrm{the} \mathrm{calorimeter} \mathrm{at} \mathrm{temperature} {\mathrm{T}}_1 = {25}^0\mathrm{C}$ is mixed with N another known mass of pure water ${\mathrm{m}}_2$ = 0.400 kg at a temperature ${\mathrm{T}}_2 = {30}^0\mathrm{C} \mathrm{and} \mathrm{final} \mathrm{temperature} \mathrm{is} \mathrm{found} \mathrm{to} \mathrm{be} \mathrm{T} = {27}^0\mathrm{C}. \mathrm{Let} {\mathrm{s}}_{\mathrm{w}}$ and W be the heat capacity of water and the water equivalent of the calorimeter.

Heat gained by (water + calorimeter) at temperature ${\mathrm{T}}_1 = \mathrm{Heat} \mathrm{lost} \mathrm{by} \mathrm{water} \mathrm{at} \mathrm{temperature} {\mathrm{T}}_2$

i.e., ${\mathrm{m}}_1{\mathrm{s}}_{\mathrm{w}}(\mathrm{T}-{\mathrm{T}}_1)+{\mathrm{Ws}}_{\mathrm{w}}(\mathrm{T}-{\mathrm{T}}_1) = {\mathrm{m}}_2{\mathrm{s}}_{\mathrm{w}}({\mathrm{T}}_2-\mathrm{T})$

     or    W = ${\mathrm{m}}_2\left(\frac{{\mathrm{T}}_2-\mathrm{T}}{\mathrm{T}-{\mathrm{T}}_1}\right)-{\mathrm{m}}_1$

Hence, W = 0.400 kg [$\frac{{30}^0\mathrm{C}-{27}^0\mathrm{C}}{{27}^0\mathrm{C}-{25}^0\mathrm{C}}$] - 0.150 kg

                 = 0.450 kg