Question 347: Properties of Matter

Question

Steam at 100oC is passed into 20 g of water at 10oC. When water acquires a temperature of 80oC, the mass of water present will be
[Take specific heat of water = 1 cal g$$-$$1 oC$$-$$1 and latent heat of steam = 540 cal g$$-$$1]

Accepted Answer

According to the principle of calorimetry.
Heat lost = Heat gained
$$m{L_v} + m{s_w}\Delta 	heta  = {m_w}{s_w}\Delta 	heta $$
$$ \Rightarrow $$ m ÃÂ 540 + m ÃÂ 1 ÃÂ (100 Ã¢ÂÂ 80)
= 20 ÃÂ 1 ÃÂ (80 Ã¢ÂÂ 10)
$$ \Rightarrow $$  m = 2.5 g
Therefore total mass of water at 80ÃÂ°C
= (20 + 2.5) g = 22.5 g

Answer

24 g

Answer

31.5 g

Answer

42.5 g

Answer

22.5 g

Properties of Matter

Steam at 100oC is passed into 20 g of water at 10oC. When water acquires a temperature of 80oC, the mass of water present will be [Take specific heat of water = 1 cal g$-1 oC-1 and latent heat of steam = 540 cal g-$1]

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