Steam at 100 °C is passed into 1 kg of water contained in a calorimeter at 9 °C till the temperature of water and calorimeter is increased to 90 °C. The mass of the steam condensed is nearly (Water equivalent of calorimeter = 0.1 kg Specific heat of water = 1 calg-1 °C-1 Latent heat of vapourisation = 540 calg-1)

Question

Steam at 100 °C is passed into 1 kg of water contained in a calorimeter at 9 °C till the temperature of water and calorimeter is increased to 90 °C. The mass of the steam condensed is nearly (Water equivalent of calorimeter = 0.1 kg Specific heat of water = 1 calg-1 °C-1 Latent heat of vapourisation = 540 calg-1) (A) 81 g (B) 162 g (C) 243 g (D) 486 g

Tardigrade Admin · Accepted Answer

Let mass of the steam condensed is x. Heat released = Heat gained by water ⇒ x × 540+x × 1 ×(100-90) =1 × 1 ×(90-9)+0.1 × 1 ×(90-9) ⇒ 540 x+10 x=81+8.1 ⇒ x=(89.1/550)=0 .162 kg , x=162 g

81 g

162 g

243 g

486 g

Let mass of the steam condensed is x. Heat released = Heat gained by water ⇒x×540+x×1×(100−90) =1×1×(90−9)+0.1×1×(90−9) ⇒540x+10x=81+8.1 ⇒x=55089.1​=0.162kg,x=162g

Let mass of the steam condensed is $x$ .
Heat released $=$ Heat gained by water
$\Rightarrow x \times 540 + x \times 1 \times (100 - 90)$
$= 1 \times 1 \times (90 - 9) + 0.1 \times 1 \times (90 - 9)$
$\Rightarrow 540 x + 10 x = 81 + 8.1$
$\Rightarrow x = \frac{89.1}{550} = 0.162 k g, x = 162 g$