$\mathrm{M}$ grams of steam at $100^{\circ} \mathrm{C}$ is mixed with $200 \mathrm{~g}$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} \mathrm{C}$ [heat of vaporization of water is $540 \mathrm{cal} / \mathrm{g}$ and heat of fusion of ice is $80 \mathrm{cal} / \mathrm{g}]$, the value of $\mathrm{M}$ is_____________
(40) Using the principal of calorimetry
$M_{\mathrm{ice}} L_{f}+m_{\mathrm{ice}}(40-0) C_{w}$
$=m_{\text {stream }} L_{v}+m_{\text {stream }}(100-40) C_{w}$
$\Rightarrow \quad M(540)+M \times 1 \times(100-40)$
$\quad=200 \times 80+200 \times 1 \times 40$
$\Rightarrow \quad 600 M=24000$
$\Rightarrow \quad M=40 g$
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