AP EAMCET · PHYSICS · Thermal Properties of Matter
Steam at \(100^{\circ} \mathrm{C}\) is passed into 114 g of water at \(30^{\circ} \mathrm{C}\). The mass of water present in the mixture when the temperature of the water becomes \(70^{\circ} \mathrm{C}\) is (Latent heat of steam \(=540 \mathrm{cal} \mathrm{g}^{-1}\); specific heat capacity of water \(=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) )
- A \(122 g\)
- B \(132 g\)
- C \(142 g\)
- D \(152 g\)
Answer & Solution
Correct Answer
(A) \(122 g\)
Step-by-step Solution
Detailed explanation
\(Q_{\text{gained}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}}\) \(Q_{\text{gained}} = 114 \text{ g} \times 1 \text{ cal g}^{-1} {^{\circ}\mathrm{C}}^{-1} \times (70 - 30) {^{\circ}\mathrm{C}} = 4560 \text{ cal}\)…
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