Question

1 g of steam at $ 100{}^\circ C $ and equal mass of ice at $ 0{}^\circ C $ are mixed. The temperature of the mixture in steady state will be (latent heat of steam $ =540\text{ }cal/g, $ latent heat of ice = 80 $ cal/g $ ):

• A. $ 50{}^\circ C $
• B. $ 100{}^\circ C $
• C. $ 67{}^\circ C $
• D. $ 33{}^\circ C $
• E. $ 0{}^\circ C $

Answer 1

Answer: (B)

Heat taken by ice to raise its temperature, to $ 100{}^\circ C $ $ {{Q}_{1}}=1\times 80+1\times 1\times 100 $ $ =180\text{ }cal $ Heat given by steam when condensed $ {{Q}_{2}}={{m}_{2}}{{L}_{2}} $ $ =1\times 540=540\text{ }cal $ As $ {{Q}_{2}}>{{Q}_{1}}, $ hence, temperature of mixture will remain $ 100{}^\circ C $ .