# If 1.80 m3 of water at 0°C is frozen and cooled to -10°C by being in contact with a great

###### Question:

I've solved the problem, but its not taking my answer. Please help if you can tell where I went wrong. Thanks!
Solve for Mi
MwLf = Mi * Ci * deltaT
Mi = 1800(79.7)/5 = 28692 kg
Change in entropy of water = Sw
-Q/T = 79.7 * 10^3 / 273 = -291.9 kcal/K
Change in entropy of ice = Si
Q/T = 28692 * .5 * 10 / 263 = 545.5 kcal/K
Total change in entropy = Si + Sw = 545 + -291.9 = 253.6 kcal/K * 4186J/kcal = 106159.6 J/K
Please let me know if you can find the mistake?

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