Question

If 10 g of ice at 0${}^o$C mixes with 10 g of water at 10${}^o$C , then the final temperature t is given by

• A. $(10\times 80)+10(t-0)=10(10-t)$
• B. $10\times 80=10(10-t)+10(t-0)$
• C. $t=5^{o}C$
• D. $t=0^{o}C$

Answer 1

The correct option is D $t=0^{o}C$

Heat released by water $=10\times 1\times 10=100cal$

Heat to be absorbed by ice $=10\times 80=800cal$

Heat released is very very less than heat absorbed so final temperature is $0^{o}C$