TS EAMCET · Physics · Thermal Properties of Matter
Steam of mass 60 g at a temperature \(100^{\circ} \mathrm{C}\) is mixed with water of mass 360 g at a temperature \(40^{\circ} \mathrm{C}\). The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is \(540 \mathrm{cal} \mathrm{g}^{-1}\) and specific heat capacity of water is \(1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) )
- A \(1: 20\)
- B \(1: 10\)
- C \(1: 5\)
- D \(1: 3\)
Answer & Solution
Correct Answer
(A) \(1: 20\)
Step-by-step Solution
Detailed explanation
For water, \(\mathrm{m}_1=360 \mathrm{~g}, \mathrm{~T}_1=40^{\circ} \mathrm{C}, \mathrm{T}_2=100^{\circ} \mathrm{C}\) \(\therefore\) Heat required, \(\mathrm{H}_1=\mathrm{m}_1 \mathrm{C}_1\left(\mathrm{~T}_2-\mathrm{T}_1\right)\)…
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