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10 g of ice at `0^@C` is mixed with 100 g of water at `50^@C`. What is the resultant temperature of mixtureA. `31.2^@C`B. `32.8^@C`C. `36.8^@C`D. `38.2^@C` |
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Answer» Correct Answer - D `100xxs_(omega)(50-theta)=10L_("ice")+10xx1xx(theta-0)` `100xx1(50-theta)=10xx80+10theta` `5000-1000=800+10theta` `1100=4200` `theta=38.2^@C` |
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