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

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