A(n) 90-g ice cube at 0°C is placed in 680 g of water at 26°C. What is the final temperature of the mixture?
Using
Heat lost = Heat gained
Let the final temperature be T
For converting 90g ice to water at 0 deg
H = mCf = 90*333 = 29970 J
Now
29970 + mCw(dT) = MC(dT)
29970 + 90*4.186(T-0) = 680*4.186(26-T)
29970 + 376.74T = 74008.48-2846.48T
3223.22T = 44038.48
T = 13.66o C
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