Asked by shekhar14351 | 19th Feb, 2010, 11:01: PM
The heat required to increase the temperature of ice to 0 °C and melt it is,
= mCpT + mλ= (5)(0.5)(10) + (5)(80) = 425 cal
The heat that can supplied by, water at 60 °C,
= mCpT = (20)(1)(60 - 0) = 1200 cal
So after providing 425 cal as required by first calculation, 1200 - 425 = 775 cal are left with the TOTAL amount of water (20gm + 5 gm of melted ice).
775 = mCpT = (25)(1)(T - 0)
T = 31 °C.
The final temperature of the mixture would be, 31 °C.
Answered by | 20th Feb, 2010, 06:19: AM
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