JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
\(M\) 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 \(42\)
Answer & Solution
Correct Answer
(C) \(40\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M} \times 540+\mathrm{M}+60=200 \times 80+200 \times 1 \times(40-0)\) \(\Rightarrow \mathrm{M}=40\)
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