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NEET PHYSICSThermal Properties of MatterMedium

Question

Steam at 100C100^{\circ}\text{C} is passed into 20 g20\text{ g} of water at 10C10^{\circ}\text{C}. When water acquires a temperature of 80C80^{\circ}\text{C}, the mass of water present will be (Take specific heat of water =1 cal g1C1= 1\text{ cal g}^{-1\circ}\text{C}^{-1} and latent heat of steam =540 cal g1= 540\text{ cal g}^{-1})

A

24 g24\text{ g}

B

31.5 g31.5\text{ g}

C

42.5 g42.5\text{ g}

D

22.5 g22.5\text{ g}

Step-by-Step Solution

According to the principle of calorimetry, Heat lost = Heat gained. Let mm be the mass of steam condensed. The heat lost by mm gram of steam at 100C100^{\circ}\text{C} to condense into water at 100C100^{\circ}\text{C} and then cool down to 80C80^{\circ}\text{C} is: Qlost=m×Lv+m×s×ΔTsteamQ_{\text{lost}} = m \times L_v + m \times s \times \Delta T_{\text{steam}} Qlost=m×540+m×1×(10080)Q_{\text{lost}} = m \times 540 + m \times 1 \times (100 - 80) Qlost=540m+20m=560mQ_{\text{lost}} = 540m + 20m = 560m

The heat gained by 20 g20\text{ g} of water at 10C10^{\circ}\text{C} to reach 80C80^{\circ}\text{C} is: Qgained=M×s×ΔTwaterQ_{\text{gained}} = M \times s \times \Delta T_{\text{water}} Qgained=20×1×(8010)=20×70=1400 calQ_{\text{gained}} = 20 \times 1 \times (80 - 10) = 20 \times 70 = 1400\text{ cal}

Equating the heat lost and heat gained: 560m=1400560m = 1400 m=1400560=2.5 gm = \frac{1400}{560} = 2.5\text{ g}

Thus, the total mass of water present at the end is the initial mass of water plus the mass of condensed steam: Total mass of water =20 g+2.5 g=22.5 g= 20\text{ g} + 2.5\text{ g} = 22.5\text{ g}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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