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10-1.Thermometry, Thermal Expansion and Calorimetry
medium
$M$ 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 $540 \;cal/\mathrm{g}$ and heat of fusion of ice is $80 \;\text { cal/g }]$ the value of $\mathrm{M}$ is
A
$35$
B
$37$
C
$40$
D
$42$
(JEE MAIN-2020)
Solution
$\mathrm{M} \times 540+\mathrm{M}+60=200 \times 80+200 \times 1 \times(40-0)$
$\Rightarrow \mathrm{M}=40$
Standard 11
Physics