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 $=1 cal g^{-1}{ }^{\circ} C^{-1}$ and latent heat of steam $=540 cal g^{-1}$ ]
(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:
- (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 $
