Properties Of Solids And Liquids Question 574
Question: 5 g of ice at $ 0{}^\circ C $ is dropped in a beaker containing 20 g of water at $ 40{}^\circ C $ . The final temperature will be
Options:
A) $ 32{}^\circ C $
B) $ 16{}^\circ C $
C) $ 8{}^\circ C $
D) $ 24{}^\circ C $
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Answer:
Correct Answer: B
Solution:
For water and ice mixing $ {\theta _{\text{mix}}}=\frac{m _{W}{\theta _{W}}+\frac{{m _{i}}{L _{f}}}{c _{W}}}{{m _{i}}+m _{W}} $
$ =\frac{20\times 40-\frac{5\times 80}{1}}{5+20}=16{}^\circ C $
BETA
