SATHEE: Thermal Properties of Matter

Thermal Properties of Matter - Result Question 18

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 $= 1 c a l g^{- 1}^{\circ} C^{- 1}$ and latent heat of steam $= 540 c a l g^{- 1}$ ]

(a) $24 g$

(b) $31.5 g$

(d) $22.5 g$

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Answer:

Correct Answer: 18. (d)

Solution:

(d) According to the principle of calorimetry.

Heat lost $=$ Heat gained

$m L_{v} + m s_{w} Δ θ = m_{w} s_{w} Δ θ$

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

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

$\Rightarrow m = 2.5 g$

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

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

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Thermal Properties of Matter - Result Question 18

18. Steam at 100∘C is passed into 20g of water at 10∘C. When water acquires a temperature of 80∘C, the mass of water present will be: [2014] [Take specific heat of water =1calg−1∘C−1 and latent heat of steam =540calg−1 ]

Answer:

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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 $= 1 c a l g^{- 1}^{\circ} C^{- 1}$ and latent heat of steam $= 540 c a l g^{- 1}$ ]