Steam is passed into `54 gm` of water at `30^(@)C` till the temperature of mixture becomes `90^(@)C`. If the latent heat of steam is `536 cal//gm`, the mass of the mixture will be

Options

A. `80 gm`
B. `60 gm`
C. `50 gm`
D. `24 gm`

Answer

B

Explanation

Let mass of steam condensed = `M gm`
`M xx 536 = mxx1xx(100-90)`
`=54xx1xx(90-30)`
`Mxx536=54xx60`
`M = (54xx60)/(536) ~=6gm`
Hence, mass of mixture = `54+6 = 60gm`.