TS EAMCET · Physics · Thermal Properties of Matter
When 54 g of ice at \(-20^{\circ} \mathrm{C}\) is mixed with 25 g of steam at \(100^{\circ} \mathrm{C}\), then the final mixture at thermal equilibrium contains
- A 20 g water at \(1.00^{\circ} \mathrm{C}\)
- B 73 g water at \(100^{\circ} \mathrm{C}\) and 6 g steam at \(100^{\circ} \mathrm{C}\)
- C 8 g steam at \(100^{\circ} \mathrm{C}\) and 12 g water at \(0^{\circ} \mathrm{C}\)
- D 20 g water at \(50^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) 73 g water at \(100^{\circ} \mathrm{C}\) and 6 g steam at \(100^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} m_{\text {ice }}=54 \mathrm{~g} & ; & T_{\text {ice }}=-20^{\circ} \mathrm{C} \\ m_{\text {steam }}=25 \mathrm{~g} & ; & T_{\text {stem }}=100^{\circ} \mathrm{C} \\ \end{array}\) # Heat Required to bring ice to \(0^{\circ} \mathrm{C}\) :-…
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