Thermal Properties of Matter - Result Question 18

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18. Steam at $100^{\circ} C$ is passed into $20 g$ of water at $10^{\circ} C$. When water acquires a temperature of $80^{\circ} C$, the mass of water present will be: [2014] [Take specific heat of water $=1 cal g^{-1}{ }^{\circ} C^{-1}$ and latent heat of steam $=540 cal g^{-1}$ ]

======= ####18. Steam at $100^{\circ} C$ is passed into $20 g$ of water at $10^{\circ} C$. When water acquires a temperature of $80^{\circ} C$, the mass of water present will be: [2014] [Take specific heat of water $=1 cal g^{-1}{ }^{\circ} C^{-1}$ and latent heat of steam $=540 cal g^{-1}$ ]

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/thermal-properties-of-matter/thermal-properties-of-matter—result-question-18.md (a) $24 g$

(b) $31.5 g$

(c) $42.5 g$

(d) $22.5 g$

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Answer:

Correct Answer: 18. (d)

Solution:

  1. (d) According to the principle of calorimetry.

Heat lost $=$ Heat gained

$mL_v+ms_w \Delta \theta=m_w s_w \Delta \theta$

$\Rightarrow m \times 540+m \times 1 \times(100-80)$

$\Rightarrow 20 \times 1 \times(80-10)$

$\Rightarrow m=2.5 g$

Therefore total mass of water at $80^{\circ} C$

$ =(20+2.5) g=22.5 g $