Questions
Steam is passed into 54 g of water at 30°C till the temperature of the mixture becomes 90°C. If the latent heat of steam is 536 cal/g, the mass of the mixture will be:
a
80 g
b
60 g
c
50 g
d
24 g
detailed solution
Correct option is B
Let mass of steam be m thenm×536+m×1×(100-90)=54×1×(90-30) ⇒ m=5.93 g≈6 g Total mass of mixture =54+6=60 g.Talk to our academic expert!
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