A piece of iron of mass 16 g is dropped (at a temperature of$112.5°\text{C}$ ) into a cavity in a block of ice. So that it melts 2.5g ice. If the latent heat of fusion of ice is=$80\text{cal/g}$ ) the specific heat of iron is

Mass of iron ${\text{m}}_{\text{iron}}=16\text{g}$

Initial temperature${\text{T}}_1=112.5°\text{C}$

When it is dropped into the cavity of ice block

${\text{m}}_{\text{ice}}\text{melted}=2.5\text{g}$

$\text{L}=80\text{cal/g}$

Temperature of ice${\text{T}}_{\text{2}}=\text{0°C}$

Final temperature of iron${\text{T}}_{\text{f}}=\text{0°C}$

We know

Heat lost by hot body = heat gained by cold body

${\text{m}}_{\text{iron}}\times {\text{s}}_{\text{iron}}\left({\text{T}}_{\text{i}}-{\text{T}}_{\text{f}}\right)={\text{m}}_{\text{ice}}\times \text{L}$

$\Rightarrow \text{16}\times {\text{s}}_{\text{iron}}\left(112.5-0\right)=2.5\times 80$

${\text{s}}_{\text{iron}}=\frac{2.5\times 80}{16\times 112.5}=0.11\text{cal/g°C}$

$=\text{0}{\text{.11calg}}^{-1}{\text{°C}}^{-1}$