M grams of steam at


$\mathrm{M}$ 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

$M_{\mathrm{ice}} L_{f}+m_{\mathrm{ice}}(40-0) C_{w}$

$=m_{\text {stream }} L_{v}+m_{\text {stream }}(100-40) C_{w}$

$\Rightarrow \quad M(540)+M \times 1 \times(100-40)$

$\quad=200 \times 80+200 \times 1 \times 40$

$\Rightarrow \quad 600 M=24000$

$\Rightarrow \quad M=40 g$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now