Question

2 g of water condenses when passed through 40 g of water initially at 25°C. the condensation of steam raises the temperature of water to 54.3°C . what is the latent heat of steam ?

A) 540cal/g
B) 536cal/g
C) 270cal/g
D) 380cal/g

solution with procedure pls.

Answer 1

answer is A
let L be the latent heat and using principle of calorimetry .2L+2(100?54.3)p L=540.3cal/gm

Answer 2

Given :

Mass of steam = 2g

Mass of water = 40g

Initial temperature of water = 25°C

Final temperature of water = 54.3°C

To Find :

The latent heat of steam

Solution :

• From the principle of calorimetry, we have the relation

          Heat lost by steam = Heat gained  by the water

          $mL+ms\Delta t=ms\Delta t$

• By substituting the values in the equation

         $(2\times L)+[2\times 1\times (100-54.3)]=40\times (54.3-25)$

         $2L = 1080.6$

         $L=540.3 cal/gm$

         $L = 540cal/gm$

The latent heat of steam is 540cal/gm