question_answer

Steam is passed into 22 gm of water at 20°C. The mass of water that will be present when the water acquires a temperature of 90°C (Latent heat of steam is 540 cal/gm) is [SCRA 1994]

A)            24.8 gm                                  

B)            24 gm

C)            36.6 gm                                  

D)            30 gm

Correct Answer: A

Solution :

                   Let m gm of steam get condensed into water (By heat loss). This happens in following two steps.                    Heat gained by water (20°C) to raise it?s temperature upto 90° \[=22\times 1\times (90-20)\]                    Hence, in equilibrium heat lost = Heat gain                    Þ \[m\times 540+m\times 1\times (100-90)=22\times 1\times (90-20)\]                    Þ \[m=2.8\]gm The net mass of the water present in the mixture \[=22+2.8=24.8\,gm.\]