The water equivalent of copper calorimeter is 4.5g, the specific heat of copper is $0.09{{\mathrm{calg}}^{-1}}^{\circ }{\mathrm{C}}^{-1}$.Among the following incorrect one is

Given that water equivalent of copper calorie meter is 4.5 g.

Water equivalent = ms in grams

So W = ms = 4.5 g

$\text{ We know specific heat of copper }=0.09\mathrm{cal}/{\mathrm{gm}}^{\circ }\mathrm{C}$

$\mathrm{m}\times 0.09=4.5\mathrm{g}$

(i)we get mass of calorimeter

$\mathrm{m}=\frac{450}{9}=50\mathrm{g}$

So option (a) is false

(ii) thermal capacity of calorimeter

$\mathrm{H}=\mathrm{ms}=4.5{\mathrm{cal}}^{\circ }{\mathrm{c}}^{-1}$

So option (b) is true

(iii) the heat required to raise the calorimeter by 8⁰c is

We know

$\begin{array}{l}\mathrm{Q}=\mathrm{ms\Delta t}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta t}=8\\ =4.5\times 8 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{ms}=4.5{\mathrm{cal}}^{\circ }{\mathrm{c}}^{-1}\end{array}$

$\mathrm{Q}=36\mathrm{cal}$

So option (c)is also true

(iv) Heat required to melt 15 g of ice in calorimeter

${\mathrm{Q}}_{\text{toel }}=\text{ heat required to melt }15\mathrm{g}\text{ of ice }+\text{ heat required for cal orim eter to be at }{0}^{\circ }\mathrm{C}$

$Q=15\times 80\mathrm{cal}+0=1200\mathrm{cal}$

So option (d) is also true.

So we get only option (a) as incorrect among the given options.