Question

Steam is passed into 22 gm of water at 20°C. The mass of water that will be present when the water acquires a temperatue of 90°C (Latent heat of steam is 540 cal/g) is(a) 24.83 gm(b) 24 gm(c) 36.6 gm(d) 30 gm

Answer 1

   ANWSER:

                      Let us consider mass of the water =m grams

                       here the following steps will happen

                       100 degrees of steam enters into 100 degrees of water

                              H1=m*540

                              H2=m*1*8(100-90)

                          90=22*1*(90-20)

                         here heat loss=heat gain

                         m*540+m*1*(100-90)=22*p m

                          by solving the above eqation you will get

                        m=22+2.8=24.8 grams