WBJEE · Physics · Thermal Properties of Matter
\(300 \mathrm{gm}\) of water at \(25^{\circ} \mathrm{C}\) is added to \(100 \mathrm{gm}\) of ice at \(0^{\circ} \mathrm{C}\). The final temperature of the mixture is
- A \(12.5^{\circ} \mathrm{C}\)
- B \(0^{\circ} \mathrm{C}\)
- C \(25^{\circ} \mathrm{C}\)
- D \(50^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(0^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Ref. at \(0^{\circ} \mathrm{C}\) water \((\Delta \mathrm{H})\) liberated \(=300 \times 1 \times 25=7500 \mathrm{cal}\) \((\Delta \mathrm{H})\) required \(=\mathrm{m}_{\text {ice }} \mathrm{L}_{\mathrm{f}}\) \(=100 \times 80 \mathrm{cal}=8000 \mathrm{cal}\) as…
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