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 _{i}}}{c _{W}}}{{m _{i}}+m _{W}} $

$ =\frac{20\times 40-\frac{5\times 80}{1}}{5+20}=16{}^\circ C $



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